# Answer on Mechanics | Relativity Question for Patrick

Question #25160

A ball of mass 0.150kg is dropped from rest from a height of 1.25m. It rebounds from the floor to reach a height of 0.960m. what impulse was given to the ball by the floor?

Expert's answer

Impulse is defined as a changein momentum. The momentum of the ball just before interacting with the floor is m*v_1, where v_1 is the velocity of the ball at hitting the floor. Let's determine this velocity. From conservation of energy, m*g*h_1 = (1/2)*m*(v_1)^2;

v_1 = sqrt(2*g*h_1);

In the similar fashion, just after rebounding from the floor, v_2 = sqrt(2*g*h_2), where h_2 is the height reached after rebounding.

If we define the direction away from the floor (up) as positive and towards the floor (down) as negative, and project the velocities on this axis,

v_1 = - sqrt(2*g*h_1);

v_2 = sqrt(2*g*h_2).

The change of momentum, taking account of the change of direction, is expressed as

dp = p2 - p1 = m*v_2 - m*v_1 = m * ( sqrt(2*g*h_1) + sqrt(2*g*h_2) ).

Substituting values, we get

dp = 0.150 kg * ( sqrt(2*(9.806 m/s^2)*(1.25 m)) + sqrt(2*(9.806 m/s^2)*(0.96

m)) ) = 1.3935 kg*m/s

Answer: impulse of about 1.4 kg*m/s

v_1 = sqrt(2*g*h_1);

In the similar fashion, just after rebounding from the floor, v_2 = sqrt(2*g*h_2), where h_2 is the height reached after rebounding.

If we define the direction away from the floor (up) as positive and towards the floor (down) as negative, and project the velocities on this axis,

v_1 = - sqrt(2*g*h_1);

v_2 = sqrt(2*g*h_2).

The change of momentum, taking account of the change of direction, is expressed as

dp = p2 - p1 = m*v_2 - m*v_1 = m * ( sqrt(2*g*h_1) + sqrt(2*g*h_2) ).

Substituting values, we get

dp = 0.150 kg * ( sqrt(2*(9.806 m/s^2)*(1.25 m)) + sqrt(2*(9.806 m/s^2)*(0.96

m)) ) = 1.3935 kg*m/s

Answer: impulse of about 1.4 kg*m/s

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