A 10.0 kg block A is released from rest a height h above a 5.00 kg plate P. whic
ID: 1290232 • Letter: A
Question
A 10.0 kg block A is released from rest a height h above a 5.00 kg plate P. which can slide freely along the smooth vertical guides BC and DE as shown in the figure. Just before hitting the plate. the block has a velocity of 6.00 m/s. The coefficient of restitution for the collision between the block and the plate is e = 0.75. The spring has a spring constant of 2.00 kN/m and before the plate and block collide the spring is compressed by 100 mm from its unstretched position. You may neglect the mass of the spring. (a) Determine the velocity of the plate just alter the collision. (b) Find the maximum Compression of the spring due to the impact.Explanation / Answer
a) e = (v2-v1)/(u1-u2) = v2-v1/6
v2 = 4.5+v1
LCM
m1*u1 + m2*u2 = m1*v1 + m2*v2
(10*6)+(5*0) = (10*v1) + (5*(4.5+v1))
4 = 10*v1 + 22.5 + 5v1
v1 = 1.23 m/s
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b) E1 = 0.5*m1*u1^2 + 0.5*k*x^2 + m2*g*0.5
E1 = (0.5*10*6*6) + (0.5*2000*0.1*0.1)+(5*9.8*0.5) = 214.5 J
E2 = 0 + 0.5*k*x1^2 + m2*g*(0.5-x1) + m1*g*(0.5-x1)
E2 = 0.5*2000*x^2 + (5*9.8*x1) +24.5 + 49 = 1000x1^2 - 49x1 + 73.5
E2 = E1
1000x1^2 + 49x1 + 73.5 = 214.5
1000x1^2 + 49x1- 141 = 0
x1 = 0.351 m = 351mm
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