A ball of mass 400 grams moves perpendicularly towards a vertical wall at a constant speed of 16 meters per second. Given that the wall exerts an impulse of 11 newton seconds on the ball during the impact, find the rebound speed of the ball.
We will begin by sketching a diagram modeling the situation before and after the impact. We are told that a ball of mass 400 grams is traveling at a speed of 16 meters per second toward a vertical wall. During the impact, the wall exerts an impulse of 11 newton seconds on the ball. This results in the ball rebounding with a speed of 𝑣 meters per second in the opposite direction. It is this speed 𝑣 that we are trying to calculate. We recall that the impulse acting on a body is equal to the momentum after the collision minus the momentum before the collision. As momentum is equal to mass multiplied by velocity, this can be written using the equation 𝐽 is equal to 𝑚𝑣 minus 𝑚𝑢, where 𝐽 represents the impulse.
In this question, we will let the positive direction be the direction the ball was moving initially. This means that the impulse and direction of motion after the collision are both negative. Before substituting in the values, we’ll convert the mass to kilograms using the fact that there are 1000 grams in one kilogram. The mass of the ball is therefore equal to 0.4 kilograms.
As already mentioned, the impulse 𝐽 is equal to negative 11. The momentum after the collision will be equal to 0.4 multiplied by negative 𝑣. And the momentum before the collision is equal to 0.4 multiplied by 16. The equation simplifies to negative 11 is equal to negative 0.4𝑣 minus 6.4. We can add 0.4𝑣 and 11 to both sides such that 0.4𝑣 is equal to 4.6. Dividing through by 0.4 gives us 𝑣 is equal to 11.5. If a ball of mass 400 grams collides with a vertical wall at a speed of 16 meters per second, where the wall exerts an impulse of 11 newton seconds on the ball, then the rebound speed is equal to 11.5 meters per second.