Question

A basketball player drops a 0.60 kg basketball vertically so that it is traveling 6.0 m/s when it reaches the floor. The ball rebounds upward at a speed of 4.2 m/s. (a) Determine the magnitude and direction of the ball’s change in momentum. (b) Determine the average net force that the floor exerts on the ball if the collision lasts 0.12s.

Answer 1

Concepts and reason

The concepts required to solve the question is generalized impulse-momentum principle.

Initially, apply the equation of change in momentum of ball to find the magnitude of the ball’s change in momentum. With the help of that, find the direction of momentum. Then, use the impulse momentum theorem to find the net force that the floor exerts on the ball.

Fundamentals

Change in momentum is mathematically given by,

$\Delta p = {p_f} - {p_i}$ …… (1)

Here, $\Delta p$ is the change in momentum, ${p_f}$ is the final momentum, and ${p_i}$ is the initial momentum.

The momentum is given by,

$p = mv$ …… (2)

Here, m is mass and v is velocity.

The relation of force exerted and the momentum is called as the impulse-momentum principle. It is given by,

$\Delta p = F\Delta t$ …… (3)

Here, F is the force, and $\Delta t$ is the change in time.

(a)

Substitute equation (2) in equation (1).

$\begin{array}{c}\\\Delta p = {p_f} - {p_i}\\\\ = m{v_f} - m{v_i}\\\\ = m\left( {{v_f} - {v_i}} \right)\\\end{array}$

Substitute 0.60 kg for m, -6.0 m/s for ${v_i}$ and 4.2 m/s for ${v_f}$.

$\begin{array}{c}\\\Delta p = m\left( {{v_f} - {v_i}} \right)\\\\ = 0.60\;{\rm{kg}}\left( {4.2\;{\rm{m/s}} - \left( { - 6.0\;{\rm{m/s}}} \right)} \right)\\\\ = 6.12\;{\rm{kg}} \cdot {\rm{m/s}}\\\end{array}$

(b)

Rearrange equation (3) to find the value of force exerted.

$F = \frac{{\Delta p}}{{\Delta t}}$

Force F is applied by surface on ball in upward direction and by ball on surface in downward direction.

$\begin{array}{c}\\\frac{{\Delta p}}{{\Delta t}} = F\\\\ = {N_s} - {F_b}\\\end{array}$

Here, N_s is the normal force applied by the surface and F_b is the force applied by the ball.

$\begin{array}{c}\\\frac{{\Delta p}}{{\Delta t}} = {N_s} - {F_b}\\\\ = {N_s} - mg\\\end{array}$

Thus,

${N_s} = \frac{{\Delta p}}{{\Delta t}} + mg$

Substitute $6.12\;{\rm{kg}} \cdot {\rm{m/s}}$ for $\Delta p$, 0.12 s for $\Delta t$, 0.60 kg for m and 9.8 m/s² for g.

$\begin{array}{c}\\{N_s} = \frac{{6.12\,{\rm{kg}} \cdot {\rm{m/s}}}}{{0.12\;{\rm{s}}}} + \left( {0.60\;{\rm{kg}}} \right)\left( {9.8\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}} \right)\\\\ = 56.9\;{\rm{N}}\\\end{array}$

The average net force that the floor exerts on the ball is 56.9 N.

Ans: Part a

The magnitude of the change in momentum is $6.12\;{\rm{kg}} \cdot {\rm{m/s}}$ and direction is positive.

Part (b)

The average net force that the floor exerts on the ball is 56.9 N.