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A 10-kg block A is released from rest 2 m above the 5-kg plate P, which can slide freely along the smooth vertical guides BC and DE. Determine the velocity of the block and plate just after impact. The coefficient of restitution between the block and the plate is e = 0.75. Also, find the maximum compression of the spring due to impact.The spring has an unstretched length of 600 mm

The spring with the 5 kg plate are stationary before impact at 450mm. The 600mm unstretch length of the spring is without the 5 kg plate

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Answer:

Check the explanation

Step-by-step explanation:

Given data

Mass of the block A is Ma = 10kg

Mass of the plate is Mp = 5kg

Coefficient of restitution between the block and that of the plate is e=0.75

Unstretched length of the spring is 0.6m

From the conservation of energy, we have four motion of block A from (1) to (2)


(Ta)_1 + (Va)_1 = (Ta)_2 + (Va)_2

Then accordingly, we will have the step by step calculation in the attached image below .

A 10-kg block A is released from rest 2 m above the 5-kg plate P, which can slide-example-1
A 10-kg block A is released from rest 2 m above the 5-kg plate P, which can slide-example-2
A 10-kg block A is released from rest 2 m above the 5-kg plate P, which can slide-example-3
A 10-kg block A is released from rest 2 m above the 5-kg plate P, which can slide-example-4
A 10-kg block A is released from rest 2 m above the 5-kg plate P, which can slide-example-5
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