# Laws of Motion

## NCERT Solution

### Part 2

Question 11: A truck starts from rest and accelerates uniformly at 2.0 m s^{-2}. At t = 10s, a stone is dropped by a person standing on the top of the truck (6 m high from the ground). What are the (a) velocity, and (b) acceleration of the stone at t = 11 s? (Neglect air resistance)

**Answer:** Velocity after 11 s can be calculated as follows:

`v=u+at`

`=0+10xx2=20` m/s

Now, we need to calculate the vertical component of velocity

`v=u+g\t`

`=0+10xx1=10` m/s

Resultant velocity can be calculated as follows:

`R^2=20^2+10^2`

= 400 + 100 = 500

Or, R = `sqrt(500)=10sqrt5`

The moment stone is dropped the horizontal net force on it becomes zero. So, net acceleration on stone at 11 s is equal to g = 10 m s^{-2}

Question 12: A bob of mass 0.1 kg hung from the ceiling of a room by a string 2 m long is set into oscillation. The speed of the bob at its mean position is 1 m s^{-1}. What is the trajectory of the bob if the string is cut when the bob is (a) at one of its extreme positions, (b) at its mean position?

**Answer:** When the bob is at one of its extreme position, its velocity is zero. So, when the string is cut the bob will fall down in a straight line which normal to the horizon.

When the bob is at mean position, its velocity is non-zero. Moreover, a downward force due to gravitational acceleration is also acting on it. Due to combined effect of horizontal and vertical components the bob will behave like a projectile.

Question 13: A man of mass 70 kg stands on a weighing scale in a lift which is moving

(a) Upwards with a uniform speed of 10 m s^{-1}

**Answer:** In each case, the weighing machine measures the reaction R which is apparent weight. As the lift is moving with uniform speed, its acceleration is zero.

R = mg = 70 × 10 = 700 N

(b) Downwards with a uniform acceleration of 5 m s^{-2}

**Answer:** R = m (g + a) = 70(10 - 5) = 70 × 5 = 350 N

(c) Upwards with a uniform acceleration of 5 m s^{-2}

**Answer:** R = m(g + a) = 70 × 15 = 1050 N

(d) What would be the reading if the lift mechanism failed and it hurtled down freely under gravity?

**Answer:** R = m (g – g) = 70 × 0 = 0 N

Question 14: This figure shows the position-time graph of a particle of mass 4 kg. What is the (a) force on the particle for t < 0, t > 4 s, 0 < t < 4s? (b) Impulse at t = 0 and t = 4 s? (Consider one-dimensional motion only)

**Answer:** When t < 0 then the particle is at rest. So, the net force on particle is zero

When t > 0, the graph shows a straight line parallel to time axis. This implies that the particle is at rest. So force on particle is zero.

When 0 < t < 4, the particle is moving at constant velocity. So, force is zero.

Impulse at t = 0

Here, u = 0, v= ¾ = 0.75 m/s, m = 4 kg

Impulse = change in momentum

`=mv-mu=4xx0.75=3` kg m s^{-1}

Impulse at t = 4 s

`=mv-mu=0-4xx0.75=-3` kg m s^{-1}

Question 15: Two bodies of masses 10 kg and 20 kg respectively kept on a smooth, horizontal surface are tied to the ends of a light string. A horizontal force F = 600 N is applied to (i) A, (ii) B along the direction of string. What is the tension in the string in each case?

**Answer:** Mass of the system = 10 + 20 = 30 kg

Acceleration can be calculated as follows:

F = ma

Or, `a=F/m`

`=(600)/(30)=20` m s^{-2}

When force is applied no A then

`F-T=m_1a`

Or, `T=F-m_1a`

`=600-10xx20=400` N

When force is applied on B then

`T=600-20xx20=200` N

Question 16: Two masses 8 kg and 12 kg are connected at the two ends of a light inextensible string that goes over a frictionless pulley. Find the acceleration of the masses, and the tension in the string when the masses are released.

**Answer:** Two masses can shown by following figure

The bigger mass will move downward with acceleration a while the smaller mass will move upward

We have, m_{1} = 8 kg, m_{2} = 12 kg and g = 10 m s^{-2}

Applying the second law of motion, equation of motion can be written as follows:

For mass m_{1}

`T-m_1g=m_1a`

For mass m_{2}

`m_2g-T=m_2a`

Adding two equations we get

`(m_2-m_1)g=(m_1+m_2)a`

Or, `(12-8)10=20a`

Or, `20a=40`

Or, `a=(40)/(20)=2` m s^{-2}

Now, tension in string can be calculated as follows:

`m_2g-T=m_2a`

Or, `T=m_2(g-a)`

`=12(10-2)=12xx8=96` N

Question 17: A nucleus is at rest in the laboratory frame of reference. Show that if it disintegrates into two smaller nuclei the products must move in opposite directions.

**Answer:** Let us assume that mass of nucleus is m and masses of two resultant nuclei are m_{1} and m_{2}

Let us assume that velocity of two nuclei are v_{1} and v_{2} after disintegration.

Initial momentum = 0 because the nucleus is at rest

As per Law of conservation of momentum

Initial momentum = final momentum

So, 0 = `m_1v_1+m_2v_2`

Or, `m_1v_1=-m_2v_2`

The negative sign means that two smaller nuclei move in opposite directions.

Question 18: 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?

**Answer:** Initial momentum of each ball p_{i} = 0.05 × 6 = 0.3 kg m s^{-1}

As the balls change their direction of motion after collision so final momentum of each ball

= p_{f} = 0.05 × -6 = - 0.3 kg m s^{-1}

Impulse = change in momentum

`=p_f-p_i`

`=-0.3-0.3=-0.6` kg m s^{-1}

Question 19: A shell of mass 0.020 kg is fired by a gun of mass 100 kg. If the muzzle speed of the shell is 80 m s^{-1}, what is the recoil speed of the gun?

**Answer:** Momentum of shell = m × v

`=0.020xx80=1.6` km m s^{-1}

Momentum of shell should be equal to that of gun

So, `1.6=100×v`

Or, `v=(1.6)/(100)=0.016` m s^{-1}

Question 20: A batsman deflects a ball by an angle of 45° without changing its initial speed which is equal to 54 km/h. What is the impulse imparted to the ball? (Mass of the ball is 0.15 kg)

**Answer:** Initial speed of ball = 54 km/h `=54xx5/(18)=15` m/s and m = 0.15 kg

Let O be the position of bat. AO shows the path of ball when it strikes the ball, OB shows the path when the ball goes after hitting the bat. Angle ∠AOB = 45°

Initial momentum of ball = mu cos θ

= 0.15 × 15 xx cos 22.5

= 0.15 × 15 × 0.9239 along NO

Final momentum of ball

= mu cos θ along ON

Impulse = change in momentum

= mu cos θ - ( - mu cos θ)

= 2 mu cos θ

= 2 × 0.15 × 15 × 0.9239 = 4.16 kg m s^{-1}