**(1) A bullet of mass 100g penetrates a distance of 10cm through a fixed block of mass 2kg. If the block is free to move, the distance penetrated will be nearly**

(a) 3.5cm (b) 4.5cm (c) 5.5cm (d) 7.5cm (e) 9.5cm

If ‘F’ is the retarding force offered by the block and ‘s’ is the distance of penetration when the block is fixed, we have

(a) 3.5cm (b) 4.5cm (c) 5.5cm (d) 7.5cm (e) 9.5cm

Fs = ½ mu^2 ……………..(1)

where ‘m’ is the mass and ‘u’ is the initial velocity of the bullet. If ‘x’ is the distance of penetration when the block is free to move, we have

Fx = ½ mu^2 - ½ (M+m) v^2 where ‘M’ is the mass of the block and ‘v’ is the common velocity of the bullet and the block after the impact. Evidently, v = mu/(M+m) from the momentum conservation law so that we have

Fx = ½ mu^2 - ½ (M+m) [mu/(M+m)]^2

Simplifying, Fx = ½ Mmu^2 /(M+m) ………….(2)

Dividing eq(2) by eq(1) we get x/s = M/(M+m) so that x = Ms/(M+m) = 2×0.1/(2+0.1) = 0.095m = 9.5cm.

**(2) An engine delivering constant power displaces a large block through 40cm in 4 seconds. What will be the displacement of the block in 16seconds?**

(a) 1.6m (b) 3.2m (c) 4.8m (d) 6.4m (e) 8m

(a) 1.6m (b) 3.2m (c) 4.8m (d) 6.4m (e) 8m

If the power is constant, the displacement will be proportional to t^(3/2) as shown below:

Power P = Fv = m(dv/dt)×v with usual notations. Since the power is constant (k) we have m(dv/dt)×v = k so that v(dv) = k(dt)/m. Integrating, we get

½ v^2 = kt/m. Therefore, v α√t. Since s=vt the displacement ‘s’ is proportional to t√t. Remember this:

**In the above problem therefore we can write**

*If power is constant, displacement is directly proportional to t^(3/2)*

0.4/x = (4√4)/(16√16) =8/64, from which x = 3.2m