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# physics Q?

block 1 of mass m1 slides from rest along a frictionless ramp from height h and then collides with stationary block 2, which has mass m2 = 3m1. After the collision, block 2 slides into a region where the coefficient of kinetic friction is μk and comes to a stop in distance d within that region. What is the value of distance d if the collision is (a) elastic and (b) completely inelastic? Express your answer in terms of the variables given and g.

(a) d =

(b) d =

### 1 Answer

- peterpanLv 72 months ago
a)

m1 v01 = m1 v1 + m2 v2

1/2 m1 v01^2 = 1/2 m1 v1^2 + 1/2 m2 v2^2

v1 = [(m1 - m2) v01 ] / (m1 + m2) = - 1/2 v01

v2 = (2 m1 v01) / (m1 + m2) = 1/2 v01

thenμk m2 g d = 1/2 m2 v2^2

d = v2^2 / (2 μk g)

v2 = 1/2 v01 = 1/2 (2 g h)^1/2

then

d = h / (4 μk)

b)

m1 v01 = (m1 + m2) v

v = (m1 v01) / (m1 + m2) = 1/4 v01

then

μk m2 g d = 1/2 m2 v^2

d = v^2 / (2 μk g)

v = 1/4 v01 = 1/4 (2 g h)^1/2

then

d = h / (16 μk)

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