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With what speed does the mass hit the ground at the bottom of the cliff?The following...
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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High School Teacher
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
Posted by nessus on July 9, 2011 at 8:13 PM (Answer #1)
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