Block A with mass of 300kg is moving on a horizontal floor towards a mass B with

Question

Block A with mass of 300kg is moving on a horizontal floor towards a mass B with mass of 100kg at rest. The coefficient of kinetic friction between the floor and the block is 0.5. When a is 15 meters away from B, it's velocity is V(0). A and B collide head_on, and after the collision, block A comes to rest at a distance of 5meters from the collision point. Block B comes to rest 20 meters away from the collision point. Using work-energy and momentum principles,

(A) find the velocities of both blocks just after the collision

(B) the velocity of A just before the collision

(C) the initial velocity V(0)

Explanation / Answer

initial velocity of A, before collision = u
final velocity of A after collision = v
final velocity of B after collision = v'
friction force = mu*mg
A) now, ke of A after collision = 0.5*300*(v)^2 = 0.5*300*9.8*5 [ from work energy theorem, work done by friction = KE of block]
v = 7 m/s
similiarly, 0.5*100*(v')^2 = 0.5*100*9.8*20
v' = 14 m/s

B) from m omentum conservation
300*u = 300*7 + 100*14
u = 11.667 m/s
c) Initial Velocity, Vo
0.5*300*Vo^2 - 0.5*300*11.667^2 = 0.5*300*9.8*15
Vo = 16.82 m/s

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