A firecracker is placed in the middle of a cluster of stationary billiard balls. The firecracker-billiard ball system sits on a frictionless surface. As the firecracker explodes, the balls scatter. Assuming no air resistance, what is the momentum of the system after the explosion of the firecracker?

a) greater than the momentum of the system before the explosion

b) equal to the combined speed of the balls

c) equal to the mass of the balls

d) zero

According to law of conservation of momentum, the momentum of the fire cracker - billiard ball system will be zero.

Answer: D

Explanation:

As we know that before the explosion the total momentum exhibited by the fire cracker and billiard board system is zero as both the objects are at rest.

As explosion is an internal force of the fire cracker, which on hitting the billiard balls will make them fall in different directions.

So the speed at which the balls will be falling at different directions will have same magnitude but different directions. And the mass of the balls will be same.

So the total momentum of the fire-cracker and billiard ball system will be zero as momentum is a vector quantity which considers the magnitude and direction of the speed of the balls.

Thus the total momentum after explosion will be zero.

Momentum is one of those things that are 'conserved' (like energy). That means the total amount of momentum in an isolated system is the same after some event as it was before the event.

Before the ear-splitting explosion, while the firecracker and the billiard balls were just sitting there on the frictionless table, the momentum of the system was zero, because nothing was moving. Right there, that tells us that the total momentum of all the balls and the fragments of the firecracker ... if you add up all the momentum vectors AFTER the explosion ... the sum is still zero. (D)