Impulse and Momentum (finding net force)

[acra]

A 0.500-kg ball is dropped from rest at a point 1.20 m above the floor. The ball reboundes straight upward to a height of 0.700 m. What are the magnitude and direction of the impulse of the net force applied to the ball during the collision with the floor?

I have been trying to solve this various ways, I have tried using PE and KE, but I keep on coming up with 3.1305 instead of the 4.28 which I should be getting.

Any guidance would be greatly appreciated.

 

[Andrew Mason]

A 0.500-kg ball is dropped from rest at a point 1.20 m above the floor. The ball reboundes straight upward to a height of 0.700 m. What are the magnitude and direction of the impulse of the net force applied to the ball during the collision with the floor?

I have been trying to solve this various ways, I have tried using PE and KE, but I keep on coming up with 3.1305 instead of the 4.28 which I should be getting.

Any guidance would be greatly appreciated.

You have to determine the ball's change in momentum. To do that you need to know its speed immediately before and immediately after the impact with the floor.

Can you determine these speeds from the two heights given?

AM

 

[kyle8921]

The impulse-momentum theorem states that the change in momentum of an object is equal to the net force applied to the object multiplied by the time interval over which it acts. In this case, we can use this theorem to find the magnitude and direction of the impulse of the net force applied to the ball during the collision with the floor.

First, we need to calculate the initial and final momentum of the ball. The ball is initially at rest, so its initial momentum is zero. When it rebounds, it will have a final velocity v, which we can calculate using the equation v^2 = u^2 + 2as, where u is the initial velocity (zero), a is the acceleration due to gravity (9.8 m/s^2), and s is the displacement (1.20 m). Solving for v, we get v = 4.38 m/s.

The final momentum of the ball can be calculated using the equation p = mv, where m is the mass of the ball (0.500 kg) and v is the final velocity (4.38 m/s). Therefore, the final momentum of the ball is 2.19 kg*m/s.

Using the impulse-momentum theorem, we can now calculate the net force applied to the ball during the collision with the floor. The change in momentum is equal to the final momentum minus the initial momentum, so we have:

Δp = 2.19 kg*m/s - 0 kg*m/s = 2.19 kg*m/s

The time interval over which this change in momentum occurs is the time it takes for the ball to go from its initial position to its final position, which is the time it takes to fall from a height of 1.20 m and then rebound back to a height of 0.700 m. This can be calculated using the equation t = √(2s/g), where s is the displacement (1.20 m) and g is the acceleration due to gravity (9.8 m/s^2). Solving for t, we get t = 0.439 s.

Now, we can calculate the magnitude of the net force applied to the ball during the collision using the equation F = Δp/t. Plugging in our values, we get:

F = 2.19 kg*m/s / 0.439 s = 4.99 N

Therefore, the magnitude of the impulse of the net force applied to