# Question Video: Studying the Collision of Two Projectiles with a Moving Body Mathematics

Two projectiles of the same mass of 50 kg were fired at 44 m/s toward a target of mass 400 kg that was moving away from the projectiles at 11 m/s. Both projectiles hit the target and sunk into it. This combined body was then hit again from the opposite direction by another projectile of mass 220 kg. Given that this body also coalesced with the target and, as a result of the impact, this new body came to rest, determine the speed of the last projectile that hit the body.

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### Video Transcript

Two projectiles of the same mass of 50 kilograms were fired at 44 meters per second toward a target of mass 400 kilograms that was moving away from the projectiles at 11 meters per second. Both projectiles hit the target and sunk into it. This combined body was then hit again from the opposite direction by another projectile of mass 220 kilograms. Given that this body also coalesced with the target and, as a result of the impact, this new body came to rest, determine the speed of the last projectile that hit the body.

In order to answer any question of this type, we need to draw before and after diagrams. Letβs consider the first collision. We are told that two projectiles of mass 50 kilograms, that is, a total mass of 100 kilograms, were fired at a speed of 44 meters per second. They were fired towards a target of mass 400 kilograms that was moving in the same direction at a speed of 11 meters per second. We are told that the projectiles sunk into the target. This means that the three particles coalesced and is an example of an inelastic collision.

We know that the conservation of momentum states that the momentum before is equal to the momentum after. As there is no energy lost in the collision, we can use the equation π one π’ one plus π two π’ two is equal to π one π£ one plus π two π£ two. The first particle, which is the sum of both projectiles, has a mass of 100 kilograms and is traveling at 44 meters per second. The second particle has a mass of 400 kilograms and is moving at 11 meters per second. After the collision, the particle has mass of 500 kilograms and its velocity or speed is unknown. The left-hand side of the equation simplifies to 4400 plus 4400. Simplifying this, we have 8800 is equal to 500π£. Dividing both sides of this equation by 500 gives us π£ is equal to 17.6. The combined particle is moving with a speed of 17.6 meters per second in the same direction.

We will now clear some space and consider the second collision. We now have a body of mass 500 kilograms moving at a speed of 17.6 meters per second. Moving towards this at an unknown speed is a body of mass 220 kilograms. We are told that after the collision, the bodies coalesce and that the combined body of mass 720 kilograms comes to rest. We can now substitute these values into the equation. Before the collision, we have 500 multiplied by 17.6. As the second body is moving towards this, its velocity is negative. Therefore, we have 220 multiplied by negative π£. This is equal to 720 multiplied by zero.

Simplifying this equation gives us 8800 minus 220π£ is equal to zero. We can add 220π£ to both sides, so this is equal to 8800. Finally, dividing both sides by 220 gives us π£ is equal to 40. The speed of the last projectile that hit the body is therefore equal to 40 meters per second.