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Collisions
Elastic vs Inelastic Collisions
A collision involves an impact between bodies where momentum is always conserved, but Kinetic Energy may or may not be conserved depending on the collision type.
$$\text{Elastic: } K.E._{initial} = K.E._{final}$$
Kinetic energy is only conserved in perfectly elastic collisions, while total energy and momentum are always conserved in all collision types.
$K.E.$=Kinetic energy(Joules (J))
$e = 1$ (perfectly elastic)
→KE fully conserved. Bodies bounce off with no energy loss.
$e = 0$ (perfectly inelastic)
→Bodies stick together. Maximum KE is lost to heat, sound, deformation.
$0 < e < 1$ (real collisions)
→Partial KE loss. Most everyday collisions fall here.
Momentum Always Conserved: Total momentum of an isolated system remains constant regardless of collision type: $m_1 v_1 + m_2 v_2 = m_1 v'_1 + m_2 v'_2$.
Energy Transformation: In inelastic collisions, KE transforms into internal energy, sound, or permanent deformation — but total energy is still conserved.
1D Elastic Collision Fundamentals
For two non-rotating bodies in a Head-on collision, the sum of their initial momenta equals the sum of their final momenta. Combined with KE conservation, this gives us enough equations to solve for both final velocities.
$$m_1 v_1 + m_2 v_2 = m_1 v'_1 + m_2 v'_2$$
Conservation of linear momentum for two colliding masses in one dimension.
$m_1, m_2$=Masses of objects 1 and 2(kg)
$v_1, v_2$=Initial velocities (before collision)(m/s)
$v'_1, v'_2$=Final velocities (after collision)(m/s)
Sign Convention: Choose a positive direction (usually right). Velocities in the opposite direction are negative.
Mass Independence: Momentum conservation applies regardless of whether the collision is elastic or inelastic.
Dimensional Check: Both sides of $m_1 v_1 + m_2 v_2 = m_1 v'_1 + m_2 v'_2$ have dimensions of $[M][L][T]^{-1}$ (kg·m/s). If your answer has different units, something went wrong.
Derivation Path: By rearranging momentum conservation as $m_1(v_1 - v'_1) = m_2(v'_2 - v_2)$ and KE conservation similarly, dividing one by the other gives the relative velocity relation.
The Relative Velocity Theorem
In a perfectly elastic collision, the Relative velocity of approach equals the Relative velocity of separation in magnitude, but reversed in direction. This is a powerful shortcut that avoids the KE conservation equation entirely.
$$v_1 - v_2 = -(v'_1 - v'_2) = v'_2 - v'_1$$
The closing speed before impact equals the separation speed after impact in elastic collisions.
$v_1 - v_2$=Relative velocity of approach(m/s)
$v'_2 - v'_1$=Relative velocity of separation(m/s)
Shortcut Power: Instead of using the quadratic KE equation, combine momentum conservation with this linear relation — much simpler algebra.
Only for Elastic: This exact equality only holds when $e = 1$. For partially inelastic collisions, the separation speed is reduced by the Coefficient of restitution: $v'_2 - v'_1 = e(v_1 - v_2)$.
Quick Verification: After solving a collision problem, check that the approach speed equals the separation speed. If it doesn't, either the collision isn't elastic or there's an error.
General Velocity Equations
Combining momentum conservation and the relative velocity relation yields closed-form expressions for each body's final velocity in terms of the masses and initial velocities.
$$v'_1 = \frac{m_1 - m_2}{m_1 + m_2} v_1 + \frac{2m_2}{m_1 + m_2} v_2$$
Final velocity of body 1 after a 1D elastic collision.
$\frac{m_1 - m_2}{m_1 + m_2}$=Mass ratio coefficient for body 1's initial velocity
$\frac{2m_2}{m_1 + m_2}$=Transfer coefficient from body 2's initial velocity
Unit Shortcut: Since masses appear as ratios $\frac{m_1 - m_2}{m_1 + m_2}$, you can use any consistent mass unit (grams, kg) — they cancel.
Symmetry: The formula for $v'_2$ has the same structure but with subscripts swapped: $v'_2 = \frac{2m_1}{m_1 + m_2} v_1 + \frac{m_2 - m_1}{m_1 + m_2} v_2$.
Dimensional Sanity: The output $v'_1$ has the same dimensions as $v_1$ and $v_2$ (m/s) since the mass ratios are dimensionless. If you accidentally mix mass units (e.g., one in grams, another in kg), the ratios are wrong.
The second velocity formula gives the final velocity of body 2, derived identically by solving the same pair of equations.
$$v'_2 = \frac{2m_1}{m_1 + m_2} v_1 + \frac{m_2 - m_1}{m_1 + m_2} v_2$$
Final velocity of body 2 after a 1D elastic collision.
$\frac{2m_1}{m_1 + m_2}$=Transfer coefficient from body 1's initial velocity
$\frac{m_2 - m_1}{m_1 + m_2}$=Mass ratio coefficient for body 2's initial velocity
When $v_2 = 0$: Simplifies to $v'_2 = \frac{2m_1}{m_1 + m_2} v_1$ — the target object always moves in the direction of the incoming object.
Verification: After calculating both $v'_1$ and $v'_2$, verify momentum conservation: $m_1 v'_1 + m_2 v'_2$ should equal $m_1 v_1 + m_2 v_2$.
Special Cases in Elastic Collisions
When two bodies of equal mass collide elastically, a remarkably simple result emerges: they exchange velocities completely.
$$m_1 = m_2 \implies v'_1 = v_2 \text{ and } v'_2 = v_1$$
Complete Swap: Each body takes the other's velocity. If one was stationary, the moving one stops and the stationary one takes off.
Newton's Cradle: This principle explains why in Newton's cradle, only the end ball swings out when the opposite end ball strikes.
Billiard Balls: A cue ball hitting a stationary target ball of equal mass stops dead — all motion transfers to the target.
When a light body collides with a much heavier stationary body, the light body bounces back with nearly unchanged speed while the heavy body barely moves.
$$m_2 \gg m_1, \; v_2 = 0 \implies v'_1 \approx -v_1, \; v'_2 \approx 0$$
Limiting case of the general elastic collision formula when the target is overwhelmingly massive.
$\frac{m_1 - m_2}{m_1 + m_2}$=Approaches $-1$ when $m_2 \gg m_1$
$\frac{2m_1}{m_1 + m_2}$=Approaches $0$ when $m_2 \gg m_1$
Wall Analogy: A ball bouncing off a wall is the extreme case — the wall (infinite mass) doesn't move at all.
Squash Player: The squash ball reverses direction off the massive front wall while the wall stays perfectly still.
Speed Preserved: Only the direction changes; the magnitude of velocity stays approximately the same.
Limiting Case: As $m_2 / m_1 \to \infty$, the result converges to exactly $v'_1 = -v_1$ and $v'_2 = 0$.
When a massive body collides with a light stationary body, the massive body continues nearly unchanged while the light body flies off at approximately twice the massive body's speed.
$$m_1 \gg m_2, \; v_2 = 0 \implies v'_1 \approx v_1, \; v'_2 \approx 2v_1$$
Limiting case of the general elastic collision formula when the projectile is overwhelmingly massive.
$\frac{m_1 - m_2}{m_1 + m_2}$=Approaches $+1$ when $m_1 \gg m_2$
$\frac{2m_1}{m_1 + m_2}$=Approaches $2$ when $m_1 \gg m_2$
Double Speed: The light object leaves at $\approx 2v_1$ — twice the heavy body's speed, not equal to it.
Truck vs Pebble: A heavy truck barely notices hitting a small stone, but the stone flies off at twice the truck's speed.
Limiting Case Check: As $m_1 / m_2 \to \infty$, the result approaches exactly $v'_2 = 2v_1$.
Coefficient of Restitution
The coefficient of restitution ($e$) measures the "bounciness" of a collision. It is the ratio of the relative velocity of separation to the relative velocity of approach.
$$e = \frac{v'_2 - v'_1}{v_1 - v_2}$$
Measures what fraction of the approach speed is retained as separation speed after collision.
$e$=Coefficient of restitution
$v'_2 - v'_1$=Relative velocity of separation(m/s)
$v_1 - v_2$=Relative velocity of approach(m/s)
$e = 1$
→Perfectly elastic — approach speed = separation speed. No KE lost.
$e = 0$
→Perfectly inelastic — bodies stick together, separation speed = 0. Maximum KE lost.
$0 < e < 1$
→Partially inelastic — some KE lost. All real-world collisions fall here.
Dimensionless: $e$ has no units — it is a pure ratio.
Material Property: $e$ depends on the materials of the colliding objects. A steel ball on marble has $e \approx 0.95$; clay on clay has $e \approx 0$.
Height Connection: For a ball dropped from height $h_1$ bouncing to height $h_2$: $e = \sqrt{\frac{h_2}{h_1}}$, since velocity at ground level $\propto \sqrt{h}$.
Proportionality: $e \propto$ separation speed for fixed approach speed — higher $e$ means faster separation.
Perfectly Inelastic Collisions
In a perfectly inelastic collision, the two bodies stick together after impact and move as one combined mass. This is the collision type that loses the maximum kinetic energy.
$$v_f = \frac{m_1 v_1 + m_2 v_2}{m_1 + m_2}$$
The combined velocity after a perfectly inelastic collision, derived from momentum conservation alone.
$v_f$=Common final velocity of the combined mass(m/s)
$m_1 + m_2$=Total combined mass after sticking(kg)
$v_2 = 0$ (stationary target)
→Simplifies to $v_f = \frac{m_1}{m_1 + m_2} v_1$ — always slower than the incoming object.
$m_1 = m_2$ and $v_2 = 0$
→$v_f = \frac{v_1}{2}$ — they move together at half the initial speed.
Maximum KE Loss: Among all collision types with the same initial conditions, perfectly inelastic loses the most kinetic energy.
Only Momentum: Since bodies stick together ($e = 0$), you cannot use KE conservation. Use momentum conservation only.
Real Examples: Car crashes (crumple zones), catching a ball (hand + ball move together), bullet embedding in a block.
Limiting Case: As $m_2 \to 0$, $v_f \to v_1$ (picking up a negligible mass barely affects you). As $m_2 \to \infty$, $v_f \to 0$ (slamming into an immovable wall — you stop).
The kinetic energy lost in a perfectly inelastic collision can be calculated directly from the masses and initial velocities without finding the final velocity first.
$$\Delta K.E. = \frac{1}{2} \frac{m_1 m_2}{m_1 + m_2} (v_1 - v_2)^2$$
The kinetic energy converted to heat, sound, and deformation in a perfectly inelastic collision.
$\Delta K.E.$=Kinetic energy lost (always positive)(Joules (J))
$\frac{m_1 m_2}{m_1 + m_2}$=Reduced mass of the system(kg)
$(v_1 - v_2)^2$=Square of relative approach speed(m²/s²)
Proportionality: $\Delta K.E. \propto (v_1 - v_2)^2$ — doubling the relative speed quadruples the energy loss.
Fraction Lost: When $v_2 = 0$, the fraction of KE lost is $\frac{m_2}{m_1 + m_2}$. A heavier target means more energy lost.
Equal Masses: When $m_1 = m_2$ and one is stationary, exactly half the KE is lost.
Limiting Case: As $m_1 \gg m_2$ (heavy hits light), fraction lost $\to 0$ — barely any energy wasted. As $m_2 \gg m_1$ (light hits heavy), fraction lost $\to 1$ — almost all KE absorbed.
Application Strategy
Worked Examples
Problem Statement
A 70 g ball moving at 9 m/s to the right hits a 140 g ball at rest. Calculate the velocities after a perfectly elastic collision.
Elastic collision velocity formulaMass ratios cancel units
Strategic Toolkit
$$v'_1 = \frac{m_1 - m_2}{m_1 + m_2} v_1, \quad v'_2 = \frac{2m_1}{m_1 + m_2} v_1$$
Step 1
Identify Variables
Extract given values. Since $v_2 = 0$, the formulas simplify. Masses appear as ratios, so grams work directly.
$m_1 = 70$ g, $v_1 = 9$ m/s, $m_2 = 140$ g, $v_2 = 0$
Step 2
Calculate v'₁
Substitute into the first velocity formula.
$v'_1 = \frac{70 - 140}{70 + 140} \times 9 = \frac{-70}{210} \times 9 = -3$ m/s (bounces back)
Step 3
Calculate v'₂
Substitute into the second velocity formula.
$v'_2 = \frac{2(70)}{70 + 140} \times 9 = \frac{140}{210} \times 9 = 6$ m/s (forward)
Step 4
Verify
Check momentum conservation: $m_1 v_1 = 70 \times 9 = 630$. After: $70(-3) + 140(6) = -210 + 840 = 630$. Confirmed.
Problem Statement
A 100 g golf ball moving at 20 m/s hits an 8 kg steel ball at rest. Compute velocities after the elastic collision.
Light-hits-heavy approximationExact vs approximate answers
Strategic Toolkit
$$v'_1 \approx -v_1, \quad v'_2 \approx \frac{2m_1}{m_1 + m_2} v_1$$
Step 1
Identify the Case
$m_2 = 8$ kg $\gg m_1 = 0.1$ kg. This is the light-hits-heavy case. We expect the golf ball to bounce back.
$m_1 = 0.1$ kg, $v_1 = 20$ m/s, $m_2 = 8$ kg, $v_2 = 0$
Step 2
Exact Calculation
Use the general formulas with actual numbers.
$v'_1 = \frac{0.1 - 8}{0.1 + 8} \times 20 = \frac{-7.9}{8.1} \times 20 \approx -19.5$ m/s
$v'_2 = \frac{2(0.1)}{8.1} \times 20 \approx 0.49$ m/s
Step 3
Compare with Approximation
The approximation gives $v'_1 \approx -20$ m/s and $v'_2 \approx 0$. Very close to exact — confirms the light-hits-heavy shortcut works well.
Problem Statement
A 1500 kg car moving at 20 m/s rear-ends a 1000 kg car moving at 10 m/s in the same direction. They lock bumpers and move together. Find the combined velocity and the kinetic energy lost.
Perfectly inelastic collisionKE loss calculation
Strategic Toolkit
$$v_f = \frac{m_1 v_1 + m_2 v_2}{m_1 + m_2}$$
Step 1
Identify Variables
Both cars move in the same direction (positive). They stick together after collision.
$m_1 = 1500$ kg, $v_1 = 20$ m/s, $m_2 = 1000$ kg, $v_2 = 10$ m/s
Step 2
Find Combined Velocity
Apply momentum conservation with a single final velocity.
$v_f = \frac{1500(20) + 1000(10)}{1500 + 1000} = \frac{30000 + 10000}{2500} = 16$ m/s
Step 3
Calculate KE Lost
Find KE before and after, then subtract.
$K.E._i = \frac{1}{2}(1500)(20^2) + \frac{1}{2}(1000)(10^2) = 300000 + 50000 = 350000$ J
$K.E._f = \frac{1}{2}(2500)(16^2) = 320000$ J
$\Delta K.E. = 350000 - 320000 = 30000$ J lost
Step 4
Verify with Formula
Cross-check using the direct KE loss formula.
$\Delta K.E. = \frac{1}{2} \cdot \frac{1500 \times 1000}{2500} \cdot (20 - 10)^2 = \frac{1}{2}(600)(100) = 30000$ J. Confirmed.
Problem Statement
A ball dropped from 2 m bounces to a height of 1.28 m. Find the coefficient of restitution between the ball and the floor.
Coefficient of restitutionHeight-based formula for e
Strategic Toolkit
$$e = \sqrt{\frac{h_2}{h_1}}$$
Step 1
Identify Variables
The ball is dropped (initial velocity = 0) from height $h_1$ and bounces to $h_2$.
$h_1 = 2$ m, $h_2 = 1.28$ m
Step 2
Apply Height Formula
For a ball dropped onto a surface, $e = \sqrt{h_2 / h_1}$ since velocity $\propto \sqrt{h}$.
$e = \sqrt{\frac{1.28}{2}} = \sqrt{0.64} = 0.8$
Step 3
Interpret
An $e$ of 0.8 means the ball retains 80% of its approach speed after each bounce. This is a fairly bouncy ball (like a tennis ball on hard court).
Problem Statement
A 2 kg ball moving at 10 m/s hits a 3 kg ball at rest. The coefficient of restitution is 0.6. Find the velocities of both balls after collision.
Coefficient of restitution combined with momentum conservationSolving two simultaneous equations
Strategic Toolkit
$$e = \frac{v'_2 - v'_1}{v_1 - v_2}, \quad m_1 v_1 + m_2 v_2 = m_1 v'_1 + m_2 v'_2$$
Step 1
Write Momentum Equation
Apply conservation of momentum.
$2(10) + 3(0) = 2v'_1 + 3v'_2 \implies 20 = 2v'_1 + 3v'_2$ ... (i)
Step 2
Write Restitution Equation
Apply $e = 0.6$ with approach speed $v_1 - v_2 = 10$.
$0.6 = \frac{v'_2 - v'_1}{10} \implies v'_2 - v'_1 = 6$ ... (ii)
Step 3
Solve Simultaneously
From (ii): $v'_2 = v'_1 + 6$. Substitute into (i).
$20 = 2v'_1 + 3(v'_1 + 6) = 5v'_1 + 18 \implies v'_1 = 0.4$ m/s
$v'_2 = 0.4 + 6 = 6.4$ m/s
Step 4
Verify Momentum
Check: $2(0.4) + 3(6.4) = 0.8 + 19.2 = 20$ kg·m/s. Matches initial momentum. Confirmed.
Memorize formulas
Final Velocity 1 (Elastic)
Velocity of body 1 after 1D elastic collision.
Formula
$v'_1 = \frac{m_1 - m_2}{m_1 + m_2} v_1 + \frac{2m_2}{m_1 + m_2} v_2$
Final Velocity 2 (Elastic)
Velocity of body 2 after 1D elastic collision.
Formula
$v'_2 = \frac{2m_1}{m_1 + m_2} v_1 + \frac{m_2 - m_1}{m_1 + m_2} v_2$
Relative Velocity Equality
Approach speed equals separation speed in elastic collisions.
Formula
$v_1 - v_2 = v'_2 - v'_1$
Coefficient of Restitution
Ratio of separation to approach speed; 1 for elastic, 0 for perfectly inelastic.
Formula
$e = \frac{v'_2 - v'_1}{v_1 - v_2}$
Perfectly Inelastic Combined Velocity
Velocity after two bodies stick together.
Formula
$v_f = \frac{m_1 v_1 + m_2 v_2}{m_1 + m_2}$
KE Loss (Perfectly Inelastic)
Energy lost when two bodies stick together.
Formula
$\Delta K.E. = \frac{1}{2} \frac{m_1 m_2}{m_1 + m_2} (v_1 - v_2)^2$
Restitution from Heights
Coefficient of restitution from bounce height ratio.
Formula
$e = \sqrt{\frac{h_2}{h_1}}$
Review
1
In an inelastic collision of two objects, which of the following quantities is NOT conserved?
Ans
Kinetic energy of the system
Review
Tap to reveal
Apply the conservation rule
In any isolated collision, total momentum is conserved, but kinetic energy is not conserved unless the collision is perfectly elastic.
Kinetic energy is the quantity that changes in an inelastic collision.
2
A 2 kg ball moving at 9 m/s collides elastically with a 4 kg ball at rest. What is the velocity of the 4 kg ball after collision?
Ans
6 m/s
Review
Tap to reveal
Use the elastic-collision transfer formula
For a target initially at rest, the final speed of the second ball is $v'_2 = \frac{2m_1}{m_1+m_2}v_1$.
$$v'_2 = \frac{2(2)}{2+4}\times 9 = 6\ \text{m/s}$$
3
Two identical balls undergo a perfectly elastic head-on collision. Ball A moves at 6 m/s and ball B is at rest. What are the final velocities of A and B respectively?
Ans
A = 0 m/s, B = 6 m/s
Review
Tap to reveal
Use the equal-mass exchange property
In a perfectly elastic head-on collision between identical masses, the moving ball stops and the stationary ball takes its speed.
$$v'_A = 0\ \text{m/s},\qquad v'_B = 6\ \text{m/s}$$
4
In a 1D elastic collision, the relative velocity of approach is 12 m/s. What is the relative velocity of separation?
Ans
12 m/s
Review
Tap to reveal
Apply the elastic relative-speed rule
For a 1D elastic collision, the speed of separation equals the speed of approach in magnitude.
$$|v'_2-v'_1| = |v_1-v_2| = 12\ \text{m/s}$$
5
A 10 kg truck moving at 5 m/s collides elastically with a stationary 0.1 kg marble. The marble's approximate velocity after collision is:
Ans
10 m/s
Review
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Use the heavy-light approximation
When a very heavy body hits a much lighter stationary object elastically, the lighter object leaves at about twice the heavy body's incoming speed.
$$v'_2 \approx 2v_1 = 2(5) = 10\ \text{m/s}$$
6
A 2 kg block moving at 10 m/s collides with a 3 kg block moving at 4 m/s in the same direction. They stick together. What is the combined velocity?
Ans
6.4 m/s
Review
Tap to reveal
Conserve momentum for a sticking collision
Because the blocks stick together, treat them as one mass after impact and divide total momentum by total mass.
$$v_f = \frac{2(10)+3(4)}{2+3} = \frac{32}{5} = 6.4\ \text{m/s}$$
7
Two objects of mass 2 kg and 3 kg are moving toward each other at 5 m/s and 10 m/s respectively. After a perfectly inelastic collision, the combined velocity is:
Ans
4 m/s in the direction of the 3 kg object
Review
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Choose a sign convention and apply momentum conservation
Take the 2 kg object's direction as positive. The total initial momentum is negative, so the combined mass must move in the 3 kg object's direction.
$$v_f = \frac{2(5)+3(-10)}{5} = -4\ \text{m/s}$$
8
A 4 kg ball moving at 6 m/s collides with a 2 kg ball at rest. They stick together. How much kinetic energy is lost?
Ans
24 J
Review
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Find the common velocity
Use momentum conservation for the stuck-together masses to get the shared final speed.
$$v_f = \frac{4(6)+2(0)}{4+2} = 4\ \text{m/s}$$
Subtract final kinetic energy from initial
Compute the initial and final kinetic energies separately, then take the difference.
$$K_i = \frac{1}{2}(4)(6^2)=72\ \text{J},\quad K_f = \frac{1}{2}(6)(4^2)=48\ \text{J},\quad \Delta K = 24\ \text{J}$$
9
A ball is dropped from 5 m and bounces back to 3.2 m. The coefficient of restitution between the ball and the floor is:
Ans
0.80
Review
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Relate bounce height to restitution
For vertical bounces, coefficient of restitution is the square root of the rebound-height ratio.
$$e = \sqrt{\frac{3.2}{5}} = \sqrt{0.64} = 0.8$$
10
A ball of mass $m$ moving at velocity $v$ collides perfectly inelastically with a stationary ball of mass $3m$. The fraction of the initial kinetic energy lost is:
Ans
3/4
Review
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Find the stuck-together speed
Use momentum conservation after the perfectly inelastic collision.
$$v_f = \frac{mv}{m+3m} = \frac{v}{4}$$
Compare kinetic energy before and after
The final kinetic energy is one-fourth of the initial kinetic energy, so the lost fraction is the remainder.
$$\frac{K_f}{K_i}=\frac{1}{4}\;\Rightarrow\;\text{fraction lost}=1-\frac{1}{4}=\frac{3}{4}$$
11
A 2 kg ball moving at 8 m/s collides head-on with a 4 kg ball moving at 2 m/s in the opposite direction. After the collision, the 2 kg ball rebounds at 4 m/s. The coefficient of restitution is:
Ans
0.8
Review
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Use momentum to find the unknown rebound speed
With the 2 kg ball reversing direction, solve the momentum equation for the 4 kg ball's final velocity.
$$2(8)+4(-2)=2(-4)+4v'_2\Rightarrow v'_2=4\ \text{m/s}$$
Apply the restitution formula
Relative speed of separation divided by relative speed of approach gives the coefficient of restitution.
$$e = \frac{4-(-4)}{8-(-2)} = \frac{8}{10} = 0.8$$
12
A ball dropped from height $h$ bounces back to $\frac{h}{4}$. The coefficient of restitution between the ball and the floor is:
Ans
0.50
Review
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Use the square-root height relation
The coefficient of restitution depends on the square root of the ratio of rebound height to drop height.
$$e = \sqrt{\frac{h/4}{h}} = \sqrt{\frac14} = \frac12 = 0.50$$
13
A 1 kg ball moving at 8 m/s collides with a 3 kg ball moving at 2 m/s in the same direction. If the collision is perfectly elastic, the velocity of the 1 kg ball after collision is:
Ans
−1 m/s
Review
Tap to reveal
Substitute into the elastic-collision formula
For a 1D elastic collision, the lighter ball's final velocity can be written directly in terms of both initial velocities and masses.
$$v'_1 = \frac{1-3}{1+3}(8) + \frac{2(3)}{1+3}(2) = -4 + 3 = -1\ \text{m/s}$$
14
A 5 kg object at 10 m/s hits a 3 kg object at rest with coefficient of restitution $e = 0.4$. The velocity of the 3 kg object after collision is:
Ans
8.75 m/s
Review
Tap to reveal
Link restitution to relative speed
The coefficient of restitution gives the post-collision relative speed, so the two final velocities differ by 4 m/s.
$$v'_2 - v'_1 = 0.4(10) = 4$$
Combine with momentum conservation
Substitute $v'_1 = v'_2 - 4$ into the momentum equation and solve for the 3 kg object's speed.
$$50 = 5(v'_2-4)+3v'_2 = 8v'_2-20 \Rightarrow v'_2 = \frac{70}{8}=8.75\ \text{m/s}$$
Question 1 of 14
0:00
1
In an inelastic collision of two objects, which of the following quantities is NOT conserved?