Answer to Question #87566 in Mechanics | Relativity for Lily Vollenbroek

Question

A 5.62 g tennis ball is loaded into 1.27 kg homemade canyon the canyon is at rest. When it is ignited, Canyon recoils a distance of 6.1 cm in .0281 seconds. Find the post explosion velocity of the Canyon/tennis ball.

Answer 1

a) We can find the post explosion velocity of the cannon from the formula:

$$v_{cannon} = \dfrac{s}{t},$$

here, $s$ is the recoil distance traveled by the cannon, $t$ is time.

Then, we get:

$$v_{cannon} = \dfrac{s}{t} = \dfrac{0.061 m}{0.0281 s} = 2.17 \dfrac{m}{s}.$$

b) We can find the post explosion velocity of the tennis ball from the law of conservation of momentum (the total initial momentum of the cannon and the tennis ball is zero, because they both are initially at rest):

$$p_{total(initial)} = p_{total(final)},$$ $$p_{total(initial)} = 0,$$ $$M_{cannon}v_{cannon} + m_{ball}v_{ball} = 0,$$ $$m_{ball}v_{ball}= -M_{cannon}v_{cannon},$$ $$v_{ball} = -\dfrac{M_{cannon}v_{cannon}}{m_{ball}}.$$

From this formula we can calculate the post explosion velocity of the tennis ball:

$$v_{ball} = -\dfrac{1.27 kg \cdot 2.17 \dfrac{m}{s} }{0.00562 kg} = -490 \dfrac{m}{s}.$$

The sign minus indicates that the post explosion velocity of the tennis ball directed in the opposite direction to the recoil velocity of the cannon.

Answer:

a) $v_{cannon} = 2.17 \dfrac{m}{s}.$

b) $v_{ball} = 490 \dfrac{m}{s}.$