Question

John has to hit a bottle with a ball to win a prize. He throws a 0.4 kg ball with a velocity of 18 m/s. It hits a 0.2 kg bottle, and the bottle flies forward at 25 m/s. How fast is the ball traveling after hitting the bottle?

Answer 1

One can solve the problem by using the law of conservation of momentum. The total momentum prior to the collision must be equivalent to the total momentum after the collision, so we have:

m1v1 + m2v2 = m1v1 + m2v2

Here, m1 is 0.4 Kg that is the mass of the ball, u1 is 18 m/s that is the initial velocity of the ball, m2 is 0.2 Kg that is the mass of the bottle, and u2 is 0 that is the initial velocity of the bottle.

v1 is the final velocity of the ball, which is to be determined, and v2 is 25 m/s that is the final velocity of the bottle.

Substituting and rearranging the equation, one can find the final velocity of the ball:

v1 = m1u1 - m2v2 / m1 = (0.4 kg) (18 m/s) - (0.2 Kg) (25 m/s) / 0.4 Kg = 5.5 m/s.

Answer 2

Explanation:

The given data is as follows.

        $m_{1}$ = 0.4 kg,     $u_{1}$ = 18 m/s

       $u_{2}$ = 0 m/s,      $v_{2}$ = 25 m/s

        $m_{2}$ = 0.2 kg,    $v_{2}$ = ?        

As we know that, according to the law of conservation of momentum

       $m_{1}u_{1} + m_{2}u_{2}$ = $m_{1}v_{1} + m_{2}V_{2}$

Hence, putting the given values into the above formula as follows.

     $m_{1}u_{1} + m_{2}u_{2}$ = $m_{1}v_{1} + m_{2}V_{2}$

$0.4 kg \times 18 m/s + 0.2 \times 0 m/s$ = $0.4 \times v_{1} + 0.2 \times 25 m/s$

          7.2 = $0.4v_{3} + 5.0$

            $v_{3}$ = 5.5 m/s

Thus, we can conclude that the ball traveling at a speed of 5.5 m/s after hitting the bottle.