In the figure, block 1 of mass m1 slides from rest along a frictionless ramp fro

Question

In the figure, block 1 of mass m1 slides from rest along a frictionless ramp from height h = 2.7 m 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 k is 0.35 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?

see the image http://edugen.wileyplus.com/edugen/courses/crs7165/art/qb/qu/c09/qu_09_G.gif

Explanation / Answer

The speed of block 1 before impact
m*g*h = 1/2*m*v^2
so v = sqrt(2*g*h) = sqrt(2*9.80*2.7) = 7.275m/s

(a) elastic (conservation of momentum) m1*7.275 = m1*v1 + 3m1*v2

or 7.275 = v1 + 3*v2 where v2 is the speed of m2 after the collision

so v1 = 7.275 - 3v2

& K is conserved also so
1/2*m1*(7.275)^2 = 1/2*m1*v1^2 + 1/2*3*m1*v2^2

or 52.92 = v1^2 + 3v2^2
now 52.92 = (7.275 - 3v2)^2 + 3v2^2
or 12v2^2 - 43.65v2 = 0

so v2 = 42.84/12 = 3.64 m/s

Now ?_k*3m*g*d = 1/2*3m*v2^2

so d = 1/2*3.64^2/(9.8*?_k) = 1/2*3.64^2/9.8*0.35) = 1.93 m

(b) the collision is completely inelastic so v2 = m1*7.275/(m1+3m1) = 1.819 m/s

Now ?_k*4m*g*d = 1/2*4m*v2^2

so d = 1/2*1.819^2/(9.8*?_k) = 1/2*1.819^2/9.8*0.35) = 0.482 m

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