College Physics by Openstax Chapter 2 Problem 56: Kinematic and Dynamic Analysis of a Steel Ball Rebounding from a Rigid Floor

Problem:

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 v_y1=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 v_y2, 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 v_y2=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*}

College Physics Chapter 2 Problems

Problem 1	Problem 2	Problem 3	Problem 4	Problem 5
Problem 6	Problem 7	Problem 8	Problem 9	Problem 10
Problem 11	Problem 12	Problem 13	Problem 14	Problem 15
Problem 16	Problem 17	Problem 18	Problem 19	Problem 20
Problem 21	Problem 22	Problem 23	Problem 24	Problem 25
Problem 26	Problem 27	Problem 28	Problem 29	Problem 30
Problem 31	Problem 32	Problem 33	Problem 34	Problem 35
Problem 36	Problem 37	Problem 38	Problem 39	Problem 40
Problem 41	Problem 42	Problem 43	Problem 44	Problem 45
Problem 46	Problem 47	Problem 48	Problem 49	Problem 50
Problem 51	Problem 52	Problem 53	Problem 54	Problem 55
Problem 56	Problem 57	Problem 58	Problem 59	Problem 60
Problem 61	Problem 62	Problem 63	Problem 64	Problem 65
Problem 66