Block A of 5 kg is released from rest on the smooth incline as shown. After trav
ID: 1320967 • Letter: B
Question
Block A of 5 kg is released from rest on the smooth incline as shown. After traveling a distance of L = 2.5 m , block A strikes block B of 10 kg , which is resting on a spring of constant k = 500 N/m. Assuming perfectly plastic impact (coefficient of restitution e = 0 ) between A and B , determine the maximum deflection of the spring after is compressed down by blocks A and B .
Block A of 5 kg is released from rest on the smooth incline as shown. After traveling a distance of L = 2.5 m , block A strikes block B of 10 kg , which is resting on a spring of constant k = 500 N/m. Assuming perfectly plastic impact (coefficient of restitution e = 0 ) between A and B , determine the maximum deflection of the spring after is compressed down by blocks A and B .Explanation / Answer
Here ,
Velocity of A before impact uA
0.5*m*uA^2 = m*g*2.5*0.6
solving for uA
uA = 5.422 m/s
Now , For a perfectly plastic collison
v*(5 + 10) = 5*5.422
v = 1.81 m/s
Using conservation of energy
0.5kx^2 = 0,5mv^2 + mgh
0.5*(5 + 10)*1.81^2 + 15*9.8 * x * 0.6= 0.5 * 500*x^2
solving for x
x = 0.724 m
the maximum deflection of the spring after is compressed down by blocks A and B is x = 0.724 m