Block A (ma=5kg) is held at rest on a curved bit of a frictionless track while b

Question

Block A (ma=5kg) is held at rest on a curved bit of a frictionless track while block B (mb=10kg) rest on a horizontal bit of the track at distance h=1.5m below the release point of Block B. When it is released from rest, block A slides down the curve and collides with block B.

a) What is Pa, the momentum of block A just before it collides with block B?

b) Block A has velocity Va=(2.4m/s) after the collision. What is the impulse imparted to block A due to its collision with block B?

c) What is the final velocity of block B after its collision with block A?

d) What is the change in Kinetic Energy of the two block system due to the collision? Based upon this answer, is the collision elastic or inelastic?

Explanation / Answer

a)
velocity of block A before collision

V1a = sqrt(2*g*h) = sqrt(2*9.8*1.5) = 5.42 m/s

P1a = m*v1a = 5*5.42 = 27.11 kg.m/s


b) impulse = chage in momentum

= m1*(v2a-v1a)

= 5*(2.4 - 5.42)

= -15.1 N.s

c) loss in momentum of block A = gain in momentum of block B


m1*(v1a - v2a) = m2*V2b

v2b = m1*(v1a-v2a)/m2

= 5*(5.42-2.4)/10

= 1.51 m/s

d) Ki = Kia + Kib = 0.5*m1*v1a^2 + 0 = 0.5*5*5.42^2 = 73.44 J

Kf = Kfa + Kfb = 0.5*m1*v2a^2 + 0.5*m2*v2b = 0.5*5*2.4^2 + 0.5*10*1.51^2 = 25.8 J

ki - kf = 47.6 J

here, kf < ki

so it is an ineleastic collision

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