Question

a 1.75 kg cannon is mounted on top of a 2.5kg stationary cart and is loaded with a 48.0 gram ball. the cannon is ignited and it launches the ball forward with a speed of 80 m/s. Determine the post-explosion velocity of the cannon and cart.

Answer 1

Answer on Question #73307 - Physics / Other

A $m_{1} = 1.75 \, \mathrm{kg}$ cannon is mounted on top of a $m_{2} = 2.5 \, \mathrm{kg}$ stationary cart and is loaded with a $m_{3} = 48.0 \, \mathrm{g}$ ball. The cannon is ignited and it launches the ball forward with a speed of $v = 80 \, \mathrm{m/s}$. Determine the post-explosion velocity of the cannon and cart.

Solution:

Let $u$ is the post-explosion velocity of the cannon and cart.

The momentum conservation law gives

$$(m_{1} + m_{2}) u + m_{3} v = 0$$

Thus

$$u = - \frac {m_{3}}{m_{1} + m_{2}} v$$ $$u = - \frac {0.048}{1.75 + 2.5} \times 80 = -0.9 \, \frac {\mathrm{m}}{\mathrm{s}}$$

Answer: $0.9 \, \frac{\mathrm{m}}{\mathrm{s}}$ backward

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