Someone please help, I am so confused with this conceptual question.

If a car mass $m$ and a truck mass $M$ travelling at the same speed $v$ collide head on and stick together, which vehicle experiences the greater magnitude of change in momentum?

I was given an explanation in terms of Newton's third law with the action and reaction force being equal and opposite so the magnitude of change in momentum is the same. I get this.

However, what I really don't get is why, if I write down the change in momentum like this ($v_2$ being final velocity of both vehicles):

Change in momentum of car = Final momentum - initial momentum = $mv_2 - mv = m(v_2-v)$

Change in momentum of truck = Final momentum - initial momentum = $Mv_2 - Mv= M(v_2-v)$

These two do not equal because the mass of the truck is larger, so wouldn't the change in momentum of the truck be greater? Why is this method of thinking about it wrong? If anyone could explain it I would be really grateful!

  • $\begingroup$ Why do you assume $V_2 - V$ is the same for the two vehicles? $\endgroup$
    – The Photon
    Commented Oct 30, 2016 at 4:59
  • $\begingroup$ Second hint: momentum is a conserved quantity. $\endgroup$
    – The Photon
    Commented Oct 30, 2016 at 4:59
  • $\begingroup$ Remember that velocity and momentum are vectors, so the direction matters. If the initial velocity of the car is $v_2$ then the initial velocity of the truck is $-v_2$. You need to include that minus sign in your equation for the momentum change. $\endgroup$ Commented Oct 30, 2016 at 5:03
  • $\begingroup$ @ThePhoton Oh! I aasumed this because in the question I was given, the vehicles stick together so I thought V2 would be the same. And the question also tells me they begin by travelling at the same speed V, thats why I thought V2-V is the same for both. Is this not correct thinking? $\endgroup$
    – melm
    Commented Oct 30, 2016 at 5:08
  • $\begingroup$ Hint 3:They may be travellling with the same speed, but if they're travelling with the same velocity, they'll never collide. $\endgroup$
    – The Photon
    Commented Oct 30, 2016 at 5:10

2 Answers 2


The two vehicles have separate momenta initially, and share a momentum after collision. While the magnitude of change in momentum for each individual vehicle can differ, the system conserves momentum as a whole.

Consider two objects colliding, one with momentum $p_1$ going right, and one going left with momentum $p_2 = -2p_1$. Before collision, it is clear that the magnitude of momentum of the second object is larger than the first, or $p_2 > p_1$.

After collision, the objects stick together and as a whole move with momentum $-p_1$, due to conservation of momentum.

So the change in $p_1$, $\Delta p_1 = -2 p_1 = p_2$

and the change in $p_2$, $\Delta p_2 = p_1$

Here, it is clear that the magnitude of change in momentum for $p_1$ is larger than the magnitude of change in momentum for $p_2$, or $|\Delta p_1| > | \Delta p_2 |$. But each momentum individually is changed by the opposing momentum.

But if we look at the momentum of the system, we see that the quantity is conserved by $p_1 + p_2 = p_1 - 2p_1 = -p_1$

This implies that although the momenta of the two objects change by different amounts, they change by amounts which will ensure that the total momentum after equals the sum of the momenta initially. $p_1$ is changed more than $p_2$, but the sum of $p_1$ & $p_2$ does not change.

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Using conservation of momentum, we know the momentum before will equal the momentum of the system after, or

$p_1 + p_2 = p_{1}'+p_{2}'$

Let $p_1$ be car momentum, $p_2$ be truck.

$p_1 = mv_1$

$p_2 = -Mv_1$

After collision, the vehicles are stuck together, so they will share their mass and velocity.

$p_{1}'+p_{2}'= v_2(m+M)$

So the conservation of momentum yields: $mv_1 - Mv_1 = v_1(m-M) = v_2(m+M)$

The momentum change in the car is equal to the momentum of the truck, and conversely, the momentum change in the truck is equal to the car.

So, $\Delta p_1 = p_2$

and, $\Delta p_2 = p_1$

Then if $|p_2|>|p_1|$, $|\Delta p_1| > |\Delta p_2|$

The change of momentum for the car is more than the change of momentum for the truck, but the changes are equal and opposite leading to the conservation of momentum.


The way you have written the change in momentum, both vehicles continue moving in the same direction after the collision as they did before. This is impossible, unless they pass through each other.

Assuming the heavier truck continues moving in the same direction, its change in momentum is $M(v_2-v)$. The car reverses direction so its change in momentum is $m(v_2+v)$.


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