**Intext Questions – 2 Page: 126,127**

**1. If action is always equal to the reaction, explain how a horse can pull a cart.**

**Solution**

When the horse walks in the forward direction (with the cart attached to it), it exerts a force in the backward direction on the Earth. An equal force in the opposite direction (forward direction) is applied on the horse by the Earth. This force moves the horse and the cart forward.

The velocity at which the horse can move by applying a force on the earth depends on the mass of the horse (and the cart attached to it). The heavier the cart, the slower the motion of the horse (for a given amount of force applied by the horse on the Earth). If the cart is too heavy, the force exerted by the horse on the Earth will be insufficient to even overcome the force of inertia. In this case, the horse will not be able to pull the cart.

**2. Explain, why is it difficult for a fireman to hold a hose, which ejects large amounts of water at a high velocity.**

**Solution**

For the hose to eject water at high velocities, a force must be applied on the water (which is usually done with the help of a pump or a motor). Now, the water applies an equal and opposite force on the hose. For the fireman to hold this hose, he must apply a force on it to overcome the force applied on the hose by the water. The higher the quantity and velocity of the water coming out of the hose, the greater the force that must be applied by the fireman to hold it steady.

**3. From a rifle of mass 4 kg, a bullet of mass 50 g is fired with an initial velocity of 35 m s ^{–1}. Calculate the initial recoil velocity of the rifle.**

**Solution**

Given, the Bullet’s mass (m_{1}) = 50 g

The rifle’s mass (m_{2}) = 4kg = 4000g

Initial velocity of the fired bullet (v_{1}) = 35 m/s

Let the recoil velocity be v_{2}.

Since the rifle was initially at rest, the initial momentum of the rifle = 0

Total momentum of the rifle and bullet after firing = m_{1}v_{1} + m_{2}v_{2}

As per the law of conservation of momentum, the total momentum of the rifle and the bullet after firing = 0 (same as initial momentum)

Therefore, m_{1}v_{1} + m_{2}v_{2} = 0

Therefore, the recoil velocity of the rifle is 0.4375 meters per second in the direction opposite to the trajectory of the bullet (backward direction).

**4. Two objects of masses 100 g and 200 g are moving along the same line and direction with velocities of 2 ms ^{–1} and 1 ms^{–1}, respectively. They collide and after the collision, the first object moves at a velocity of 1.67 ms^{–1}. Determine the velocity of the second object.**

**Solution**

Assuming that the first object is object A and the second one is object B, it is given that:

Mass of A (m_{1}) = 100g

Mass of B (m_{2}) = 200g

Initial velocity of A (u_{1}) = 2 m/s

Initial velocity of B (u_{2}) = 1 m/s

Final velocity of A (v_{1}) = 1.67 m/s

Final velocity of B (v_{2}) =?

Total initial momentum = Initial momentum of A + initial momentum of B

= m_{1}u_{1} + m_{2}u_{2}

= (100g) × (2m/s) + (200g) × (1m/s) = 400 g.m.sec^{-1}

As per the law of conservation of momentum, the total momentum before collision must be equal to the total momentum post collision.

v_{2} = 1.165 m/s

Therefore, the velocity of object B after the collision is 1.165 meters per second.