# With what speed does the mass hit the ground at the bottom of the cliff?The following takes place at a colony on the Moon some time in the distant future:A 40 kg cannon is loaded with a 10 kg...

With what speed does the mass hit the ground at the bottom of the cliff?

The following takes place at a colony on the Moon some time in the distant future:

A 40 kg cannon is loaded with a 10 kg cannonball and mounted on a 6 kg platform with wheels very low to the ground. The wheeled platform is pushed and moves forward at 3 m/s horizontally. When the cannonball is fired out, the cannon recoils (backward) with a velocity of 2 m/s relative to the ground. The cannonball then strikes a large solid block that weighs 32N and is sitting on the edge off a 50 m high cliff and becomes embedded in it. The combined mass is shot horizontally off the cliff.

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Let m1 = mass cannon+platform =46kg.

m2 = mass ball = 10kg

u1 = u2 = initial horizontal velocity of m1+m2 = 3ms^-1

v1 = final velocity of cannon and platform = -2ms^-1

Use **m1u1 + m2u2 = m1v1 + m2v2** to find m2v2 the horizontal momentum of the ball after firing. Rearranging:

m2v2 = m1u1 + m2u2 - m1v1

= (46 x 3) + (10 x 3) - (46 x (-2)) = 260kgms^-1

Now let m1u1 be the same horizontal momentum when it hits the block.

m1u1 + m2u2 = m1v1 + m2v2 note that v1 = v2 which is the final horizontal velocity of ball and block combined. Rearrange for v1:

v1 = (m1u1 + m2u2)/(m1 + m2)

= 260 + 0(stationary block)/10 + (32/1.62 (mass block))

= 8.74ms^-1 this will be the horizontal component of velocity at impact at the base of the cliff.

Find vertical component of velocity at impact. Use v^2 = u^2 +2as.

v = (u^2 + 2as)^0.5 = (0 + 2(1.62)(50))^0.5 = 12.73ms^-1

The impact speed will be the magnitude of the horizontal and vertical velocity vectors added using Pythagorus' theorem:

Impact speed = ( 8.74^2 + 12.73^2)^0.5

**= 15.44ms^-1 **