Question

A block of mass m = 0.1 kg is connected to a spring of unknown spring constant k. It is compressed to a distance x from its equilibrium position and released from rest. After approaching half the distance \left ( \frac{x}{2} \right ) from equilibrium position, it hits another block and comes to rest momentarily, while the other block moves with a velocity 3 ms -1. The total initial energy of the spring is :

Answer 1

Answer:

A block of mass m=0.1 kg is connected to a spring of unknown spring constant k. It is compressed to a distance x from its equilibrium position and released from rest. After approaching half the distance (x/2) from equilibrium position, it hits another block and comes to rest momentarily, while the other block moves with a velocity 3 ms-1. The total initial energy of the spring is:

1.5 J

0.6 J

0.3 J

0.8 J

Solution

Apply principle of conservation of momentum and energy

Momentum before collision = momentum after collision

0.1u + m*0 = 0.1*0 + m*3

½*0.1*u2 = ½*m*32

Solving these two equations, u = 3

½kx2 = ½k(x/2)2 + ½*0.1*32

3/4 kx2 = 0.9

½kx2 = 0.6 J

The correct option is B.

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