 # A ball of mass 0.150 kg is dropped from rest Sam Longoria 2021-12-18 Answered
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 0.960 m. What impulse was given to the ball by the floor?
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Step 1
In this problem, a ball of mass $m=0.150m$ is dropped from a rest height of ${h}_{1}=1.25m$. If rebounds from the floor to reach of ${h}_{2}=0.960m$. We calculate the impulse given to the ball by the floor. We use $g=9.80\frac{m}{{s}^{2}}$
Step 2
Let the upward direction be positive. Initially, the ball moves downwards, so the initial momentum is negative. Using conservation of energy, we have
$K{E}_{i}+P{E}_{i}=K{E}_{f}+P{E}_{f}$
$0+mg{h}_{1}=\frac{m{v}_{1}^{2}}{2}+0$
$mg{h}_{1}=\frac{{m}^{2}{v}_{1}^{2}}{2m}$
$mg{h}_{1}=\frac{{p}_{1}^{2}}{2m}$
${p}_{1}^{2}=2{m}^{2}g{h}_{1}$
${p}_{1}=\sqrt{2{m}^{2}g{h}_{1}}$
${\stackrel{\to }{p}}_{1}=-\sqrt{2{m}^{2}g{h}_{1}}$
Step 3
Similarly, after the rebound, the momentum must be
${\stackrel{\to }{p}}_{2}=+\sqrt{2{m}^{2}g{h}_{2}}$
Step 4
The impulse must be
$\stackrel{\to }{I}={\stackrel{\to }{p}}_{2}-{\stackrel{\to }{p}}_{2}$
$=+\sqrt{2{m}^{2}g{h}_{2}}-\left(-\sqrt{2{m}^{2}g{h}_{1}}\right)$
$=\sqrt{2{m}^{2}g{h}_{2}}+\sqrt{2{m}^{2}g{h}_{1}}$
$=\sqrt{2{m}^{2}g}\left(\sqrt{{h}_{2}}+\sqrt{{h}_{1}}\right)$
$=\left[\sqrt{2{\left(0.150kg\right)}^{2}\left(9.80\frac{m}{{s}^{2}}\right)}\right]\left(\sqrt{0.960m}+\sqrt{1.25m}\right)$
$=+1.39312N×s$
$\stackrel{\to }{I}=+1.39N×s$
Since the impulse is positive, it must be upwards.

We have step-by-step solutions for your answer! Becky Harrison
When an object receives a force for a short time, this force with time is called the impulse, so it is the area under the curve of the force-versus-time graph and it is the same as the change in momentum. The impulse is the quantity I and it is given by equation (6.4) in the form
1) $I=F\mathrm{\Delta }t=m{v}_{f}-m{v}_{i}$
When the ball is dropped from a height of 1.25 m, it gains velocity due to the gravitational acceleration. This is the initial velocity before it hits the floor. From the kinematic equations, we can calculate this velocity by
2) ${v}_{i}^{2}={v}_{0}^{2}+2g{h}_{0}$
${v}_{i}=\sqrt{0+2g{h}_{0}}$
The height is ${h}_{0}=1.25m$ Use this value into equation (2) to get the velocity of the ball before it drops on the floor
${v}_{i}=\sqrt{2g{h}_{0}}=\sqrt{2\left(9.8\frac{m}{{s}^{2}}\right)\left(1.25m\right)}=4.95\frac{m}{s}$
When the ball hits the floor, it changes its velocity. The final velocity could be calculated using equation (2) but for height, $h=0.960m$
$-{v}_{f}=-\sqrt{2gh}=-\sqrt{2\left(9.8m{s}^{2}\right)\left(0.960m\right)}=-4.34\frac{m}{s}$
Now, we plug the values for and ${v}_{i}$ into equation (1)to get the impulse
$I=m{v}_{f}-m{v}_{i}$
$=\left(0.150kg\right)\left(-4.34\frac{m}{s}-4.95\frac{m}{s}\right)$
$=-1.39kg×\frac{m}{s}$
The negative sign indicates that the direction of the impulse is upward.

We have step-by-step solutions for your answer! nick1337

Step 1
By definition, the impulse given to the ball is equal to the change of its momentum:
$J=\mathrm{\Delta }p=m\left({v}_{2}-{v}_{1}\right)$
where $m=0.15kg$ is the mass of the ball, are the projections of the ball's velocity on the vertical axis before and after the collision respectively.
Let's direct the vertical axis upwnward.
According to the mechanical energy conservation, the kinetic energy of the ball right before the collision is equal to its potential energy at the begining of the motion. Writting down the corresponding formulas for these energies, obtian:
$\frac{m{v}_{1}^{2}}{2}=mg{h}_{1}$
where $g=9.81m/{s}^{2}$ is the gravitational acceleration. and ${h}_{1}=1.25m$ is the initial height. Thus, obtain:
${v}_{1}=-\sqrt{2g{h}_{1}}$
(negative, since before the collision the velocity is directed downward).
After the collision, the kinetic energy of the ball is equal to its potential energy at the maximum height ${h}_{2}=0.96m$. Obtain:
$\frac{m{v}_{2}^{2}}{2}=mg{h}_{2}$
${v}_{2}=\sqrt{2g{h}_{2}}$
(positive, since after the collision the velocity is directed upward).
Finally, obtain the expression for the impulse:
$J=m\left({v}_{2}-{v}_{1}\right)=m\sqrt{2g}\left(\sqrt{{h}_{1}}+\sqrt{{h}_{1}}\right)$
$J=0.15×\sqrt{2×9.18}×\left(\sqrt{1.25}+\sqrt{0.96}\right)\approx 1.39N×s$

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