Question

You drop a 14-g ball from a height of 1.5 m and it only bounces back to a height of 0.85 m. what was the total impulse on the ball when it hit the floor? (ignore air resistance.)

Answer 1

The total impulse on the ball can be calculated using the principle of conservation of momentum. First, calculate the initial momentum of the ball before it hits the floor. Then, determine the time the ball is in contact with the floor. Finally, calculate the total impulse using the change in momentum and the time of contact.

Step-by-step explanation:

The total impulse on the ball when it hits the floor can be calculated using the principle of conservation of momentum. Impulse is defined as the change in momentum and is equal to the force applied multiplied by the time it is applied. In this case, since the ball is dropped and then bounces back, the change in momentum is equal to the initial momentum of the ball multiplied by -1 (due to the change in direction) and the time it is in contact with the floor.

First, we need to calculate the initial momentum of the ball before it hits the floor. Momentum is defined as the product of mass and velocity, so the initial momentum is equal to the mass of the ball (14 g = 0.014 kg) multiplied by its initial velocity.

Next, we need to determine the time the ball is in contact with the floor. This can be calculated by dividing the total height the ball bounces back (1.5 m - 0.85 m = 0.65 m) by the velocity just before it hits the floor. Since the ball is dropped, its initial velocity is 0 m/s.

Finally, we can calculate the total impulse by multiplying the change in momentum by the time of contact.

Answer 2

Remember that impulse = change in momentum

this means we compute the momentum of the ball just before impression and just after; we know the mass, so we find the speeds

the ball falls for 1.5m and will achieve a speed given by energy conservation:

1/2 mv^2 = mgh => v=sqrt[2gh]=5.42m/s

since it rises only to 0.85 m, we compute the initial speed after power from the same equation and get
v(after)=sqrt[2*9.81m/s/s*0.85m] = 4.0837 m...

now, recall that momentum is a vector, so that the momentum down has one sign and the momentum up has a positive sign, so we have

impulse = delta (mv) = m delta v = 0.014 kx (4.08m/s - (-5.42m/s) = 0.133 kgm/s