# 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).

Getting to your specific questions...

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

Remember their is only a particular object's impulse, just like you can't have a force without an object. Although the total momentum of the system does not change, since it is conserved, clearly the momentum of the individual objects changes. This means an impulse has been delivered to both of them.

Is it possible not to have an impulse during a collision?

If you define a collision as objects changing speeds, then momentum must have changed so impulses must have been delivered in every collision.

Impulse is applied over time by the normal force when two bodies are in contact. Since there is only one particle after the collision, it is hard to say there is an impulse theoretically because there would not be another body in the system for an equal and opposite force. If you think about it in a realistic context, if there is no normal force pushing the particles away from each other that would stop them combining(like how you cant punch though every wall), then yes there is no impulse. However, there is likely to be some resisitive force pushing them away from each other since things generally don't want to be combined, so you may measure an impulse in real life.