Two people are standing on a crate at rest on a frictionless horizontal surface.

Question

Two people are standing on a crate at rest on a frictionless horizontal surface. Person A has a mass of 60 kg, person B has a mass of 40 kg, the crate has a mass of 20 kg.

a) imagine the two jump off the crate together, with a speed of 3 m/s relative to the crate. What is the speed of the crate for an observer standing still on the horizontal surface.

b)imagine person A jumps before person B, each still with the speed of 3 m/s relative to the crate. What is the speed of the crate after Person A has jumped off, and what is the speed after both A and B have jumped off the crate? (Speed measured by a stationary observer)

c)imagine the two jump together, but in opposite direction. Make sure to tell me who jumps foward or backward. What is the speed of the cart now, and , if applicable, the direction?

Please explain answers!Thanks!!

Explanation / Answer

a) velocity of person be x

velocity of crate = 3-x

conservation of lineart momentum

m1v1 +m2v2 = m3v3

60*x +40*x = 20*(3-x)

100x = 60 - 20x

x = 60/120 = 0.5 m/s

velocity of crate = 3-0.5 = 2.5 m/s

b) let person A jumps with speed x

m1x = (m2+m3)*(3-x)

60*x = 60(3-x)

120x = 180

x = 180/120 = 1.5 m/s

so velocity of crate after A jumped = 1.5 m/s

now B jumps with velocity = y

velocity of crate = (3+1.5 - y)

m2*y = m3*(4.5-y)

40*y = 20(4.5-y)

60*y = 90

y = 1.5 m/s

velocity of crate = 4.5-y = 4.5-1.5 = 3m/s

c) velocity of crate = x

velocity of A = velovity of B = (3-x)

60*(3-x) - 40*(3-x) = 20*x

20(3-x) = 20*x

2x = 3

x = 1.5 m/s

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