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A.2 · 11 · A.2 Forces and momentum · conservation of momentum, Newton’s third law, elastic and inelastic collisions

Exam questions: collisions and Newton’s third law

Two collision problems, worked step by step with our own animated diagram and a recreated momentum-time graph.

SL+HLHarder26 minFree
Hook

Two balls collide. Momentum is always conserved — but kinetic energy usually is not. These two problems show what that difference looks like, ball by ball, millisecond by millisecond.

Problem 1 · one ball stops, the other moves off
1 / 4

A 0.24 kg ball stops a 0.48 kg ball

Ball X (mass 0.240 kg) moves at 16 m s⁻¹ in a straight line on a frictionless surface. It collides with a stationary ball Y (mass 0.480 kg). After the collision, ball X is stationary.

GivenValue
Mass of X0.240 kg
Mass of Y0.480 kg
Speed of X before16 m s⁻¹
Speed of Y before0 (stationary)
Speed of X after0 (stationary)
Simulation

Watch the collision, and both momentum-time lines

The idea: X arrives, they touch for a brief instant (shaded), then X is stationary and Y moves off. The two momentum lines below always add up to the same total.

Predict: since X ends up stationary, ball Y must leave with ALL of X’s original momentum. Y’s mass is exactly double X’s, so Y’s speed afterwards is…

Answer the prediction above to unlock the simulation. Then test it.

WORKED EXAMPLE · FIND Y’S SPEED, AND THE KINETIC ENERGY LOST

Ball X (0.240 kg, 16 m s⁻¹) collides with stationary ball Y (0.480 kg) on a frictionless surface; afterwards X is stationary. Find Y’s speed v, and the change in total kinetic energy.

    Check 1 of 7

    Ball X (0.24 kg, 16 m s⁻¹) hits stationary ball Y (0.48 kg). After the collision X is at rest. What is Y’s speed?

    Check 2 of 7

    In that same collision, the kinetic energy of the system…

    Check 3 of 7

    The collision lasts 2.0 ms and brings ball X (0.24 kg, 16 m s⁻¹) to rest. The magnitude of the force on X is…

    Check 4 of 7

    The force ball Y exerts on ball X, compared with the force ball X exerts on ball Y, is…

    Check 5 of 7

    Block X (mass m, speed 5v) collides with a stationary block Y and they stick together, moving off at v. What is m_Y / m_X?

    Check 6 of 7

    For that same sticking collision, the ratio (KE after) / (KE before) is…

    Check 7 of 7

    On the momentum–time graph, from t = 0 to t = 20 ms (before contact), the force on block X is…

    Apply

    A 0.50 kg trolley moving at 6.0 m s⁻¹ collides with a stationary 1.0 kg trolley and they stick together. Find their common speed afterwards, and the ratio of kinetic energy after to kinetic energy before (compare it with the 1/5 result above — is it the same collision shape?).

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    Remember earlier lessons
    From lesson A.2 · 5

    Momentum is conserved for a system when…

    From lesson A.2 · 6

    A collision where the objects stick together is called…

    Summary card

    Key points

    • • Momentum is conserved in every collision (zero net external force); kinetic energy usually is not.
    • • "They don’t stick together" does not mean "elastic" — always check the KE numbers.
    • • Impulse (F×t) equals the change in momentum of ONE object; the force pair on the two colliding objects is always equal and opposite (Newton’s third law), at every instant.
    • • For X (speed 5v) sticking to a stationary Y: m_Y/m_X = 4 and KE_after/KE_before = 1/5 — both follow from momentum conservation alone, for any mass or speed.

    Formulas

    • p_i = p_f (momentum conservation)
    • F = Δp / t
    • KE = ½mv²

    Key terms

    Perfectly inelastic collision
    :
    the colliding objects stick together and move with a common velocity; the biggest possible kinetic-energy loss for the given momentum
    Impulse
    :
    force × time, equal to the change in momentum it causes

    Common errors

    • • Assuming a collision is elastic just because nothing sticks together.
    • • Using the ORIGINAL speed instead of the SQUARED speed when computing kinetic energy.
    • • Forgetting that the two collision forces (on each object) are equal and opposite at every instant, not just at the start or end.
    Checks solved: 0 of 7
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