A soft tennis ball is dropped onto a hard floor from a height of 1.25 m and rebounds to a height of 1.05 m?

Question

A soft tennis ball is dropped onto a hard floor from a height of 1.25 m and rebounds to a height of 1.05 m?

A: Calculate its acceleration during contact with the floor if that contact lasts 3.50 ms (3.50×10-3 s) m/s.

B: How much did the ball compress during its collision with the floor, assuming the floor is absolutely rigid in m?

2 Answers

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Answer 1 · 2017-11-06T22:29:11

The ball accelerates at a rate of $- 1414 \frac{m}{s^{2}}$ $-1414m/s^2$ and compresses by $8.66 m m$ $8.66 mm$ .

Explanation:

First, we have to determine the speed the ball has at the moment it contacts the floor. Since it fell from a height of 1.25 m, the speed is found from

$v^{2} = v_{o}^{2} + 2 a d$ $v^2=v_o^2+2ad$

where $v_{o}$ $v_o$ is the starting speed (zero) , $a$ $a$ is the acceleration of the ball ( $9.8 \frac{m}{s^{2}}$ $9.8 m/s^2$ ) and $d$ $d$ is the height it falls through.

$v^{2} = 0 + 2 (9.8) (1.25) = 24.5$ $v^2 = 0 + 2(9.8)(1.25) = 24.5$

so, $v = 4.95 \frac{m}{s}$ $v = 4.95 m/s$

(By the way, this equation of motion is commonly used when we do not know the time of the motion, and do not wish to bother with calculating it!)

Next, we find the acceleration of the ball while in contact with the floor. This time, $v = 0$ $v=0$ as the final speed is zero when the ball has stopped.

$v = v_{0} + a t$ $v=v_0+at$

$0 = 4.95 + a (3.5 \times 10^{- 3})$ $0=4.95+a(3.5xx10^(-3))$

So, the acceleration is

$a = \frac{4.95}{3.5 \times 10^{- 3}} = 1414 \frac{m}{s^{2}}$ $a=4.95/(3.5xx10^(-3)) = 1414 m/s^2$

Finally, to find the deformation of the ball, we do a calculation like the first one

$v^{2} = v_{o}^{2} + 2 a d$ $v^2=v_o^2+2ad$

$0 = {4.95}^{2} + 2 (- 1414) d$ $0 = 4.95^2 +2(-1414)d$

Note that to avoid a sign conflict, I have placed a negative sign in front of the acceleration, as the ball is slowing down.

$d = \frac{{4.95}^{2}}{2828} = 0.00866 m$ $d=4.95^2/(2828) = 0.00866 m$ or 8.66 mm

Answer 2 · 2017-11-06T22:41:50

Use velocity and displacement equations for the rebound to find initial velocity. Then use this initial velocity at the time of the collision as a final velocity. Answer is:

$a = 1309.31 \frac{m}{s^{2}}$ $a=1309.31 m/s^2$

Explanation:

Question A

After the rebound, the motion on the vertical axis is deceleration, with $a = g$ $a=g$ . Therefore, the velocity $u$ $u$ and height $h$ $h$ will be equal to:

$u = u_{0} - g t$ $u=u_0-g t$

$h = u_{0} \cdot t - \frac{1}{2} g t^{2}$ $h=u_0*t-1/2g t^2$

Now, since it reaches 1.05 m, you have $h = 1.05 m$ $h=1.05m$ and $u = 0$ $u=0$ the velocity equation:

$u = u_{0} - g t$ $u=u_0-g t$
$0 = u_{0} - g t$ $0=u_0-g t$
$g t = u_{0}$ $g t=u_0$
$t = \frac{u_{0}}{g}$ $t=u_0/g$

Substituting into the height equation:

$h = u_{0} \cdot t - \frac{1}{2} g t^{2}$ $h=u_0*t-1/2g t^2$
$1.05 = u_{0} (\frac{u_{0}}{g}) - \frac{1}{2} g \cdot {(\frac{u_{0}}{g})}^{2}$ $1.05=u_0(u_0/g)-1/2g*(u_0/g)^2$
$1.05 = \frac{{(u_{0})}_{0}^{2}}{g} - \frac{1}{2} g \cdot \frac{{(u_{0})}_{0}^{2}}{g^{2}}$ $1.05=(u_0)^2/g-1/2g*(u_0)^2/g^2$
$1.05 = \frac{{(u_{0})}_{0}^{2}}{g} - \frac{1}{2} \frac{{(u_{0})}_{0}^{2}}{g}$ $1.05=(u_0)^2/g-1/2(u_0)^2/g$

The $\frac{{(u_{0})}_{0}^{2}}{g}$ $(u_0)^2/g$ is a common factor:

$1.05 = \frac{{(u_{0})}_{0}^{2}}{g} \cdot (1 - \frac{1}{2})$ $1.05=(u_0)^2/g*(1-1/2)$
$1.05 = \frac{{(u_{0})}_{0}^{2}}{g} \cdot \frac{1}{2}$ $1.05=(u_0)^2/g*1/2$
$2 \cdot 1.05 \cdot g = {(u_{0})}_{0}^{2}$ $2*1.05*g=(u_0)^2$
$u_{0} = \sqrt{2 \cdot 1.05 \cdot g}$ $u_0=sqrt(2*1.05*g)$
$u_{0} = \sqrt{2.1 g}$ $u_0=sqrt(2.1g)$

For educational purposes, $g = 10 \frac{m}{s^{2}}$ $g=10 m/s^2$ is frequently used (its actual value is $9.806 \frac{m}{s^{2}}$ $9.806 m/s^2$ . Therefore:

$u_{0} = \sqrt{2.1 \cdot 10} = \sqrt{21} ≅ 4.583 \frac{m}{s} = u_{1}$ $u_0=sqrt(2.1*10)=sqrt(21)~=4.583 m/s=u_1$ *

*This velocity will be defined as $u_{1}$ $u_1$

After its collision, the ball had an acceleration from $u = 0$ $u=0$ to $u = u_{1}$ $u=u_1$ over time $0.0035 s$ $0.0035 s$ . Therefore:

$u_{1} = u_{0} - a t$ $u_1=u_0-at$
$a t = u_{1} - u_{0}$ $at=u_1-u_0$
$a = \frac{u_{1} - u_{0}}{t}$ $a=(u_1-u_0)/t$
$a = \frac{4.583 - 0}{0.0035}$ $a=(4.583-0)/0.0035$
$a = 1309.31 \frac{m}{s^{2}}$ $a=1309.31 m/s^2$ (Answer to question A)

I actually have no way to answer question B. Either additional data is missing or it's a hard one. I assume it has to do with the energy difference before and after the collision and some formula for inelastic collisions. The total energy difference is measured between the two heights as dynamic difference:

$U_{1} - U_{2} = m \cdot g \cdot h_{1} - m \cdot g \cdot h_{2} = m \cdot g \cdot (h_{1} - h_{2}) =$ $U_1-U_2=m*g*h_1-m*g*h_2=m*g*(h_1-h_2)=$
$= m \cdot 10 \cdot (1.25 - 1.05) = 2 \cdot m$ $=m*10*(1.25-1.05)=2*m$

Hope you find it useful.

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A soft tennis ball is dropped onto a hard floor from a height of 1.25 m and rebounds to a height of 1.05 m?

A: Calculate its acceleration during contact with the floor if that contact lasts 3.50 ms (3.50×10-3 s) m/s.

B: How much did the ball compress during its collision with the floor, assuming the floor is absolutely rigid in m?

2 Answers

Explanation:

Explanation:

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Impact of this question

Science

Math

Humanities

... and beyond

A soft tennis ball is dropped onto a hard floor from a height of 1.25 m and rebounds to a height of 1.05 m?

A: Calculate its acceleration during contact with the floor if that contact lasts 3.50 ms (3.50×10-3 s) m/s. B: How much did the ball compress during its collision with the floor, assuming the floor is absolutely rigid in m?

2 Answers

Explanation:

Explanation:

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Impact of this question

A: Calculate its acceleration during contact with the floor if that contact lasts 3.50 ms (3.50×10-3 s) m/s.

B: How much did the ball compress during its collision with the floor, assuming the floor is absolutely rigid in m?