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?

Express your answer using two significant figures.

Answer 1

Answer:

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

Answer 2

Answer:

$v_{f2}$ =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}$