Question 457

Solution

Let *m*, *m*_{1}, and *m*_{2} be the respective masses of the parent nucleus and the two daughter nuclei.

The parent nucleus is at rest.

Initial momentum of the system (parent nucleus) = 0The parent nucleus is at rest.

Let

Total linear momentum of the system after disintegration = m

According to the law of conservation of momentum,

Total initial momentum = Total final momentum

0 = m

Here, the negative sign indicates that the fragments of the parent nucleus move in directions opposite to each other.

Question 458

Two billiard balls each of mass 0.05 kg moving in opposite directions with speed 6 m s^{–1} collide and rebound with the same speed. What is the impulse imparted to each ball due to the other?

Solution

Mass of each ball, m = 0.05 kg

Initial velocity of the ball = 6 m/s

Magnitude of the initial momentum of the ball, p_{1} = 0.05 x 6

= 0.3 kg m/s

After collision, the balls change their directions of motion without changing the magnitudes of their velocity.

Final momentum of each ball, p_{f }= - 0.3 kg m/s

Impulse imparted to each ball = p_{f} - p_{i} = -0.3 -0.3 = -0.6 kg m/s

The negative sign indicates that the impulses imparted to the balls are opposite in direction.

Question 459

Solution

Given,

Mass of the gun,*M* = 100 kg

Mass of the shell, Mass of the gun,

Muzzle speed of the shell,

Recoil speed of the gun =

Both the gun and the shell are at rest initially.

Initial momentum of the system = 0

Final momentum of the system =

Here, the negative sign appears because the directions of the shell and the gun are opposite to each other.

Now, as per the law of conservation of momentum. we have

Final momentum = Initial momentum

∴

= 0.020 × 80 / (100 × 1000)

= 0.016 m/s

Question 460

Solution

The given question can be illustrated as in the following fig. below.

Here,

AO = incident path of the ball,

OB = Path followed by ball after deflection,

< AOB = Angle between the incident and deflected paths of the ball = 45^{o}

Therefore, ^{}∠AOP = ∠BOP = 22.5° = *θ*

Initial velocity of the ball = Final velocity of the ball = v

On resolving the component of velocity along v, we have

Horizontal component of the initial velocity = *v*cos θ, along RO

Vertical component of the initial velocity = *v*sin θ, along PO

Horizontal component of the final velocity = *v*cos θ, along OS

Vertical component of the final velocity = *v *sin θ, along OP

The horizontal components of velocities suffer no change.

The vertical components of velocities are in the opposite directions.

So, change in linear momentum of the ball gives us the impulse which is imparted to the ball.

That is, *Impulse = mvCos*θ - (-*mvCos*θ)

= 2*mvCos*θ

Mass of the ball, m = 0.15 kg

Velocity of the ball, *v* = 54 km/h

= 15 m/s

Therefore,

Impulse = 2 x 0.15 x 15 cos 22.5^{o }= 4.16 kg m/s