# Is the impulse 0 in a perfectly inelastic collision?

I have this problem:

An object $$A$$ of mass $$48\: \text{kg}$$ hits with a velocity of $$7.0 \:\text{ms}^{-1}$$ another object $$B$$ which is not moving. After the collision the two objects move together at the same velocity of $$1.4\: \text{ms}^{-1}$$. What's the mass of $$B$$? What's the impulse applied on $$B$$ during the collision?

So this is my doubt: the mass of $$B$$ is $$192 \:\text{kg}$$, since the impulse equals the change of momentum, is the impulse equal to $$0$$ since the change of momentum doesn't change in an inelastic collision? Is it possible not to have an impulse during a collision?

• Focusing just on object B, doesn't its momentum change? – Not_Einstein May 14 '20 at 13:39
• Think of momentum as the quantity needed to completely stop a moving object. In this framework impulse = momentum for an inelastic contact when measured from a reference point co-moving with the center of mass. – John Alexiou May 15 '20 at 16:05

The momenta of individual objects in a collision do change (no matter whether it's elastic or inelastic). However, the total momentum is conserved (does not change), again, irrespective of the fact that the collision is elastic or inelastic. Thus there is a non zero, and in fact equal and opposite impulse on both the objects. Due to this equal and opposite impulse (caused by the normal force between them), the net impulse on the system of those two objects becomes zero and thus the net momentum of the system is conserved.

The initial momentum of $$A$$ is $$p_a = 48 \times 7 = 336\,\text{kgms}^{-1}$$, and the initial momentum of $$B$$ is zero since it is not moving.

Since momentum is always conserved, the total initial momentum, $$336\,\text{kgms}^{-1}$$, must equal the total final momentum, $$1.4m_a + 1.4m_b$$.

I.e

$$336 = 1.4m_a + 1.4m_b$$

So

$$m_b = \frac{336 - 1.4 \times 48}{1.4} = 192\,\text{kg}$$

as you found.

We can now calculate the impulse of $$A$$ and $$B$$.

$$\text{Imp}_a = \Delta p = 1.4\times48 - 336 = -268.8\,\text{kgms}^{-1}$$ $$\text{Imp}_b = \Delta p = 1.4\times192 - 0 = 268.8\,\text{kgms}^{-1}$$

As you can see $$\text{Imp}_b = -\text{Imp}_a$$.

This can be explained by Newton's Third Law: the force exerted on $$B$$ by $$A$$ is equal and opposite to the force exerted on $$A$$ by $$B$$ during the collision. And since they collide for the same time the impuse, $$\int Fdt$$, has the same magnitude for both objects, but different directions (signs).

Is the impulse equal to $$0$$ since the change of momentum doesn't change in an inelastic collision?