## At the county fair, Chris throws a 0.12kg baseball at a 2.4kg wooden milk bottle, hoping to knock it off its stand and win a prize. The ball

Question

At the county fair, Chris throws a 0.12kg baseball at a 2.4kg wooden milk bottle, hoping to knock it off its stand and win a prize. The ball bounces straight back at 30% of its incoming speed, knocking the bottle straight forward.

What is the bottle’s speed, as a percentage of the ball’s incoming speed?

in progress 0
1 year 2021-08-13T22:37:31+00:00 2 Answers 701 views 0

$$v_{f2}$$ =6.5%$$v_{i1}$$

Explanation:

Mass of the ball: $$m_{1} =0.12kg$$]

Initial velocity of the ball:   $$v_{i1}$$

final velocity of the ball: $$v_{f1}$$ which is -30/100 of $$v_{i1}$$ =$$-0.3v_{i1}$$

Mass of the bottle: $$m_{2} =2.4kg$$

Initial velocity of the bottle: $$v_{i2}=0m/s$$

final velocity of the bottle: $$v_{f2}$$ is unknown (to find)

by using conservation momentum, which stated that the initial momentum is equal to the final momentum.

$$m_{1} v_{i1} +m_{2} v_{i2} =m_{1} v_{f1} +m_{2} v_{f2}$$

so since the bottle is at rest firstly, therefore $$v_{i2} =0$$

$$m_{1} v_{i1} +m_{2} (0) =m_{1} v_{f1} +m_{2} v_{f2}$$

$$m_{1} v_{i1} =m_{1} v_{f1} +m_{2} v_{f2}$$         equation 1

so now substitute $$v_{f1}$$ into equation 1

$$m_{1} v_{i1} =m_{1} (-0.3v_{i1} ) +m_{2} v_{f2}$$

$$m_{1} v_{i1} = -0.3m_{1}v_{i1} +m_{2} v_{f2}$$

collect the like terms

$$m_{1} v_{i1} +0.3m_{1}v_{i1} =m_{2} v_{f2}$$

$$1.3m_{1} v_{i1} =m_{2} v_{f2}$$

divide both  side by $$m_{2}$$

$$v_{f2}=\frac{1.3m_{1} v_{i1}}{m_{2} }$$

Now substitute

$$v_{f2} =\frac{1.3*0.12*v_{i1}}{2.4}\\v_{f2} =\frac{0.156v_{i1} }{2.4} \\v_{f2} =0.065v_{i1}$$

$$v_{f2} =$$6.5%$$v_{i1}$$

v2 = 6.5% of the ball’s initial velocity

Explanation:

Let the mass of the ball and bottle be m1 and m2 respectively.

m1 = 0.12kg , m2 = 2.4kg

Let the initial and final velocity of ball be u1 and

v1 = -0.3u1 while that of the bottle be u2 and v2

The bottle was initially at rest so u2 = 0m/s

From the principle of conservation of momentum, momentum before collision is equal to momentum after collision.

m1u1 + m2u2 = m1v1 + m2v2

0.12u1 + m2×0 = 0.12(–0.3u1) + 2.4v2

0.12u1 = –0.036u1 + 2.4v2

Collecting like terms

2.4v2 = 0.12u1 + 0.036u1

2.4v2 = 0.156u1

v2 = 0.156u1/2.4

v2 = 0.065u1 = 6.5%u1