## A neutron travelling with a velocity v and K.E. ‘E’ collides perfectly elastically head on with nucleus…

Q: A neutron travelling with a velocity v and K.E. ‘E’ collides perfectly elastically head on with nucleus of an atom of mass number A at rest. The fraction of total energy retained by the neutron is :

(a) $\displaystyle (\frac{A-1}{A+1})^2$

(b) $\displaystyle (\frac{A-1}{A})^2$

(c) $\displaystyle (\frac{A+1}{A-1})^2$

(d) $\displaystyle (\frac{A-1}{A})^2$

Ans: (a)

velocity of body after collision is

$\displaystyle v_1 = \frac{(m_1-m_2)u_1}{m_1 +m_2}$

$\displaystyle v_1 = \frac{(1-A)u_1}{1 + A}$

Fraction of K.E retained is

$\displaystyle \frac{1/2 m v_1^2}{1/2 m u_1^2} = (\frac{v_1}{u_1})^2$

$\displaystyle = (\frac{(1-A)}{1 + A})^2$

## A bullet is fired at a target with a velocity of 80 m/s and penetrates 50 cm into it. If this bullet were fired into…

A bullet is fired at a target with a velocity of 80 m/s and penetrates 50 cm into it. If this bullet were fired into a target 25 cm thick with equal velocity, with what velocity would it emerge, supposing the resistance is uniform and same in both the cases?

(a) √80 m/s

(b) 40/√2 m/s

(c) 40 m/s

(d) 40√2 m/s

Ans: (d)

Sol:

Applying , $\displaystyle v^2 = u^2 + 2 a S$

$\displaystyle 0 = 80^2 + 2 a (0.5)$

a = – 6400 m⁄s2

Again , $\displaystyle v^2 = u^2 + 2 a S$

$\displaystyle v^2 = 80^2 + 2 (-6400) (0.25)$

$\displaystyle v^2 = 6400 - 3200 = 3200$

$\displaystyle v = \sqrt{3200} = 40 \sqrt{2} m/s$

## A child on a swing is 1 metre above the ground at the lowest point and 2 metre above the ground…

Q: A child on a swing is 1 metre above the ground at the lowest point and 2 metre above the ground at the highest point. The horizontal speed of the child at the lowest point of the swing is (g = 10 m/s2 ) :

(a) 4.5 m/sec

(b) 6.4 m/sec

(c) 7.8 m/sec

(d) 10 m/sec

Ans: (a)

Sol: $\displaystyle v = \sqrt{2 g (h_2 - h_1)}$

$\displaystyle v = \sqrt{2 \times 10 (2 - 1)}$

$\displaystyle v = \sqrt{20} = 4.5 m/s$

## A ball is allowed to fall down with initial speed v from a height of 10 m. It loses 50% kinetic energy…

Q: A ball is allowed to fall down with initial speed v from a height of 10 m. It loses 50% kinetic energy after striking the floor. It reaches to the same height after collision. What is the value of v ?

(a) 28 m/s

(b) 7 m/s

(c) 14 m/s

(d) it is never possible

Ans: (c)

Sol:

To reach same height of 10 m after collision,

Velocity After collision $\displaystyle = \sqrt{2 g h}$

$\displaystyle = \sqrt{2 \times 10 \times 10} = 10\sqrt{2} m/s$

As energy lost in collision is 50% , velocity becomes v⁄√2

∴ Velocity of ball on striking,

v = (10 √2) √2 = 20 m/s

Applying $\displaystyle v^2 = u^2 + 2 a S$

$\displaystyle 20^2 = u^2 + 2 \times 10 \times 10$

$\displaystyle u^2 = 20^2 - 200 = 200$

$\displaystyle u = \sqrt{200} = 10\sqrt{2}$

u = 14.14 m/s

## A ball of mass ‘m’ moving with a horizontal velocity ‘v’ strikes the bob of mass ‘m’ of a pendulum at rest. ..

Q: A ball of mass ‘m’ moving with a horizontal velocity ‘v’ strikes the bob of mass ‘m’ of a pendulum at rest. During this collision, the ball sticks with the bob of the pendulum. The height to which the combined mass raises is (g = acceleration due to gravity)

(a) $\displaystyle \frac{v^2}{4 g}$

(b) $\displaystyle \frac{v^2}{8 g}$

(c) $\displaystyle \frac{v^2}{ g}$

(d) $\displaystyle \frac{v^2}{2 g}$

Ans: (b)

Sol: Applying conservation of momentum ,

m v = (m + m)v’

v’ = v/2

Applying conservation of Energy ,

$\displaystyle \frac{1}{2} (2 m) v'^2 = (2 m) g h$

$\displaystyle h = \frac{v'^2}{2 g}$

$\displaystyle h = (\frac{v}{2})^2 \times \frac{1}{2 g}$

$\displaystyle h = \frac{v^2}{8 g}$