Question

Old naval ships fired 10 kg cannon balls from a 240 kg cannon. It was very important to stop the recoil of the cannon, since otherwise the heavy cannon would go careening across the deck of the ship. In one design, a large spring with spring constant 2.0×104 N/m was placed behind the cannon. The other end of the spring braced against a post that was firmly anchored to the ship's frame.
What was the speed of the cannon ball if the spring compressed 55 cm when the cannon was fired?

Express your answer to two significant figures and include the appropriate units.

Answer 1

Answer:

120 m/s

Explanation:

First of all, we can find the speed of the cannon recoiling. In fact, when the cannon recoils with speed V', it compresses the spring and all its kinetic energy is converted into elastic potential energy of the spring. So we can write:

$\frac{1}{2}MV'^2 = \frac{1}{2}kx^2$

where

M = 240 kg is the mass of the cannon

V is the speed of recoil of the cannon

k = 2.0×10^4 N/m is the spring constant

x = 55 cm = 0.55 m is the compression of the spring

Solving for V',

$V' = \sqrt{\frac{kx^2}{M}}=\sqrt{\frac{(2.0\cdot 10^4)(0.55)^2}{240}}=5.0 m/s$

Now we can apply instead the law of conservation of momentum to the cannon+cannonball system to find the velocity at which the cannonball is fired. Since the system is initially at rest, the initial momentum is zero, so we can write:

0=mv+MV'

where

m = 10 kg is the mass of the cannon ball

v is the velocity of the cannon ball

M = 240 kg is the mass of the cannon

V' = 5.0 m/s is the velocity of recoil of the cannon

Solving for v,

$v=-\frac{MV'}{m}=-\frac{(240)(5.0)}{10}=-120 m/s$

where the negative sign means the direction of the cannonball is opposite to that of the cannon. So, the cannonball is fired at a speed of 120 m/s.