Numerical Problems Based on Class 11 Physics Conservation of Linear Momentum

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Law of conservation of linear momentum

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The second and third laws of motion lead to one of the most important and fundamental principles of physics, called the law of conservation of linear momentum.

It can be stated as follows:

When no external force acts on a system of several interacting particles, the total linear momentum of the system is conserved. The total linear momentum is the vector sum of the linear momenta of all the particles of the system.

Numerical Problems on Conservation of Linear Momentum

Class 11 Physics · Laws of Motion

Linear Momentum Practice Set

Apply the principle $m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2$

Ch 5 · Laws of Motion

A 40 kg shell is flying at a speed of $72\text{ km h}^{-1}$. It explodes into two pieces. One of the two pieces of mass 15 kg stops. Calculate the speed of the other.

Answer $32\text{ m s}^{-1}$ 📝

Detailed Solution

According to the law of conservation of linear momentum, the total momentum before the explosion equals the total momentum after the explosion.

Initial mass $M = 40\text{ kg}$. Initial velocity $u = 72\text{ km/h} = 72 \times \frac{5}{18} = 20\text{ m/s}$.
Mass of the first piece $m_1 = 15\text{ kg}$, its velocity $v_1 = 0\text{ m/s}$.
Mass of the second piece $m_2 = 40 – 15 = 25\text{ kg}$, its velocity $= v_2$.

$$Mu = m_1v_1 + m_2v_2$$

$$40 \times 20 = (15 \times 0) + 25v_2$$

$$800 = 25v_2 \implies v_2 = \frac{800}{25} = 32\text{ m s}^{-1}$$

A bullet of mass 7 g is fired into a block of metal weighing 7 kg. The block is free to move. After the impact, the velocity of the bullet and the block is $70\text{ cm s}^{-1}$. What is the initial velocity of the bullet?

Answer $700.7\text{ m s}^{-1}$ 📝

Detailed Solution

This is a perfectly inelastic collision where the bullet embeds into the block. They move together with a common velocity $v$.

Mass of bullet $m_1 = 7\text{ g} = 0.007\text{ kg}$. Initial velocity $= u_1$.
Mass of block $m_2 = 7\text{ kg}$. Initial velocity $u_2 = 0\text{ m/s}$.
Common final velocity $v = 70\text{ cm/s} = 0.7\text{ m/s}$.

$$m_1u_1 + m_2u_2 = (m_1 + m_2)v$$

$$0.007 \times u_1 + 0 = (0.007 + 7) \times 0.7$$

$$0.007u_1 = 7.007 \times 0.7 = 4.9049$$

$$u_1 = \frac{4.9049}{0.007} = 700.7\text{ m s}^{-1}$$

A man weighing 60 kg runs along the rails with a velocity of $18\text{ km h}^{-1}$ and jumps into a car of mass 1 quintal standing on the rails. Calculate the velocity with which the car will start travelling along the rails.

Answer $1.875\text{ m s}^{-1}$ (or $1.88\text{ m s}^{-1}$) 📝

Detailed Solution

Convert the man’s velocity to m/s: $u_1 = 18\text{ km/h} = 18 \times \frac{5}{18} = 5\text{ m/s}$.

Mass of the man $m_1 = 60\text{ kg}$.
Mass of the car $m_2 = 1\text{ quintal} = 100\text{ kg}$. Initial velocity $u_2 = 0\text{ m/s}$.

Applying the principle of conservation of momentum for their combined motion:

$$m_1u_1 + m_2u_2 = (m_1 + m_2)v$$

$$60 \times 5 + 0 = (60 + 100)v$$

$$300 = 160v \implies v = \frac{300}{160} = 1.875\text{ m s}^{-1}$$

A machine gun of mass 10 kg fires 20 g bullets at the rate of 10 bullets per second with a speed of $500\text{ m s}^{-1}$. What force is required to hold the gun in position?

Answer $100\text{ N}$ 📝

Detailed Solution

By Newton’s Second Law, the force exerted by the gun on the bullets (and thus the recoil force needed to hold the gun) is equal to the rate of change of momentum of the bullets.

Mass of one bullet $m = 20\text{ g} = 0.02\text{ kg}$.
Velocity of the bullet $v = 500\text{ m/s}$.
Number of bullets fired per second $n = 10$.

Momentum of a single bullet $p = mv = 0.02 \times 500 = 10\text{ kg m s}^{-1}$.

$$F = \frac{dp}{dt} = n \times p$$

$$F = 10 \times 10 = 100\text{ N}$$

A truck of mass $2 \times 10^4\text{ kg}$ travelling at $0.5\text{ m s}^{-1}$ collides with another truck of half its mass moving in the opposite direction with a velocity of $0.4\text{ m s}^{-1}$. If the trucks couple automatically on collision, calculate the common velocity with which they move.

Answer $0.2\text{ m s}^{-1}$ 📝

Detailed Solution

Let the direction of the first truck be positive.

Mass of 1st truck $m_1 = 2 \times 10^4\text{ kg}$. Velocity $u_1 = +0.5\text{ m/s}$.
Mass of 2nd truck $m_2 = \frac{1}{2}m_1 = 1 \times 10^4\text{ kg}$. Velocity $u_2 = -0.4\text{ m/s}$ (opposite direction).

Since they couple, they move together with a common velocity $v$:

$$m_1u_1 + m_2u_2 = (m_1 + m_2)v$$

$$(2 \times 10^4 \times 0.5) + (1 \times 10^4 \times -0.4) = (2 \times 10^4 + 1 \times 10^4)v$$

$$10000 – 4000 = 30000v$$

$$6000 = 30000v \implies v = \frac{6000}{30000} = 0.2\text{ m s}^{-1}$$

Since the result is positive, the coupled trucks move in the direction of the first truck.

Show Answer

Answer: 0.2 ms^-1

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