Question

A ball of mass 0.150 kg is dropped from rest from a height of 1.25 m. It rebounds from the floor to reach a height of 0.665 m. What impulse was given to the ball by the floor?

Answer 1

Answer:

Impulse is 1.239 kg.m/s in upward direction

Explanation:

Taking upward motion as positive and downward motion as negative.

Downward motion:

Given:

Mass of ball (m) = 0.150 kg

Displacement of ball (S) = -1.25 m

Initial velocity (u) = 0 m/s

Acceleration is due to gravity (g) = -9.8 m/s²

Using equation of motion, we have:

$v_d^2=u^2+2aS\\\\v=\pm\sqrt{u^2+2aS}\\\\v_d=\pm\sqrt{0+2\times -9.8\times -1.25}\\\\v_d=\pm\sqrt{24.5}=\pm4.95\ m/s$

Since, the motion is downward, final velocity must be negative. So,

$v_d=-4.95\ m/s$

Upward motion:

Given:

Displacement of ball (S) = 0.665 m

Initial velocity ($v_d$) = 4.95 m/s(Upward direction)

Acceleration is due to gravity (g) = -9.8 m/s²

Using equation of motion, we have:

$v_{up}^2=v_d^2+2aS\\\\v_{up}=\pm\sqrt{v_d^2+2aS}\\\\v_{up}=\pm\sqrt{24.5+2\times -9.8\times 0.665}\\\\v_{up}=\pm\sqrt{10.966}=\pm3.31\ m/s$

Since, the motion is upward, final velocity must be positive. So,

$v_{up}=3.31\ m/s$

Now, impulse is equal to change in momentum. So,

Impulse = Final momentum - Initial momentum

$J=m(v_{up}-v_d)\\\\J=(0.150\ kg)(3.31-(-4.95))\ m/s\\\\J=0.150\ kg\times 8.26\ m/s\\\\J=1.239\ Ns$

Therefore, the impulse given to the ball by the floor is 1.239 kg.m/s in upward direction.