A steel ball is dropped onto a hard floor from a height of 1.50 m and rebounds to a height of 1.45 m. (a) Calculate its velocity just before it strikes the floor. (b) Calculate its velocity just after it leaves the floor on its way back up. (c) Calculate its acceleration during contact with the floor if that contact lasts 0.0800 ms (8.00×10−5 s) . (d) How much did the ball compress during its collision with the floor, assuming the floor is absolutely rigid?
Solution:
Part A
For this part, we shall consider Figure A.
We will be considering the two positions as shown. The first position is when the ball is dropped from a height of 1.50 meters. For this position, we know that y1=1.50 m, t1=0 s, and vy1=0 m/s.
Position 2 is immediately after the ball hits the floor. For this position, we do not the time elapse and the velocity but we know that the height is zero. That is y2=0 m.
Position 1 is the initial position and position 2 is the final position. Solving for vy2, we have
\begin{align*}
\left( v_{y_2} \right)^2 & = \left( v_{y_1} \right)^2 + 2 a \Delta y \\
\left( v_{y_2} \right)^2 & = \left( v_{y_1} \right)^2 + 2 a \left( y_2 - y_1 \right) \\
v_{y_2} & = \sqrt{\left( v_{y_1} \right)^2 + 2 a \left( y_2 - y_1 \right)} \\
v_{y_2} & = - \sqrt{\left( 0 \right)^2+2\left( -9.81\ \text{m/s}^2 \right)\left( 0\ \text{m} - 1.50\ \text{m} \right)} \\
v_{y_2} & = - 5.42 \ \text{m/s} \qquad {\color{DarkOrange} \left( \text{Answer} \right)}
\end{align*}\therefore The steel ball has a velocity of about 5.42 m/s directed downward when it strikes the floor.
Part B
Figure B shows the two positions we are interested in to solve for this part.
Position 1 is at the floor immediately just after the ball hits it. At this initial position, we have y1=0 m, and t1= 0 s. We do not know the velocity at this point.
At position 2, the ball bounced back to its second peak. We know that at the peak of a free falling body, the velocity is zero. So, for this final position, we have y2=1.45 m, and vy2=0 m/s. We do not know the time at this position.
Position 1 is the initial position while position 2 is the final position. Solving for the initial velocity we have
\begin{align*}
\left( v_{y_2} \right)^2 & = \left( v_{y_1} \right)^2 + 2 a \Delta y \\
\left( v_{y_2} \right)^2 & = \left( v_{y_1} \right)^2 + 2 a \left( y_2 - y_1 \right) \\
\left( v_{y_1} \right)^2 & = \left( v_{y_2} \right)^2 - 2 a \left( y_2 - y_1 \right) \\
v_{y_1} & = \sqrt{\left( v_{y_2} \right)^2 - 2 a \left( y_2 - y_1 \right)} \\
v_{y_1} & = \sqrt{\left( 0 \ \text{m/s} \right)^2 -2\left( -9.81 \ \text{m/s}^2 \right) \left( 1.45 \ \text{m} - 0 \ \text{m} \right) }\\
v_{y_1} & = 5.33 \ \text{m/s} \qquad {\color{DarkOrange} \left( \text{Answer} \right)}
\end{align*}\therefore The steel ball has a velocity of about 5.33 m/s directed upward immediately after it leaves the floor.
Part C
From our answers in Part A and Part B, we have a change in velocity from -5.42 m/s to 5.33 m/s. So, in this case the initial velocity is v1=-5.42 m/s and the final velocity is v2=5.33 m/s. We can compute for the acceleration:
\begin{align*}
a & = \frac{\Delta v}{\Delta t} \\
a & = \frac{v_2-v_1}{\Delta t} \\
a & = \frac{5.33 \ \text{m/s} - \left( -5.42 \ \text{m/s} \right)}{8.00 \times 10^{-5} \ \text{s}} \\
a & = 134,375 \ \text{m/s}^2 \\
a & = 1.34 \times 10^{5} \text{m/s}^2 \ \qquad {\color{DarkOrange} \left( \text{Answer} \right)}\\
\end{align*}Part D
The period of compression happens when the ball has a velocity of -5.42 m/s until it reaches 0 m/s. We shall solve for the change in displacement for this two given velocities. The initial velocity is -5.42 m/s and the final velocity is 0 m/s. The acceleration during this period is the one solved in Part C, a=1.34×105 m/s2.
\begin{align*}
\left( v_{y_2} \right)^2 & = \left( v_{y_1} \right)^2 + 2 a \Delta y \\
\Delta y & = \frac{\left( v_{y_2} \right)^2 - \left( v_{y_1} \right)^2}{2 a} \\
\Delta y & = \frac{\left( 0 \ \text{m/s} \right)^2 - \left( -5.42 \ \text{m/s} \right)^2}{2\left( 1.34 \times 10^5 \ \text{m/s}^2 \right)}\\
\Delta y & = -0.000110 \ \text{m} \\
\Delta y & = -1.10 \times 10^{-4} \ \text{m} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}\\
\end{align*}