From a rifle of mass 4 kg , a bullet of mass 50 g is fired with an initial velocity of 35 m s . Calculate the initial recoil velocity of the rifle.

Given data, Mass of rifle, m r = 4 kg Mass of bullet, m b = 50 g Recoil velocity of the rifle, v r = ? The initial velocity of the bullet, v b = 35 m s We know that initially the rifle and bullet are at rest. Therefore, the initial momentum (momentum before firing) of the rifle bullet system will be zero. P 1 = 0 The total momentum of the rifle bullet system after firing, P 2 = ( m r × v r ) + ( m b × v b ) ⇒ P 2 = ( 4 × v r ) + ( 50 1000 × 35 ) = 4 v r + 1 . 75 Now, according to the law of conservation of momentum, total momentum of the system before firing = total momentum of the system after firing, i.e., P 1 = P 2 ⇒ 0 = 4 v r + 1 . 75 ⇒ v r = - 0 . 4375 m s The negative sign indicates that the rifle recoils backwards with a velocity of 0 . 4375 m s

Two objects of masses 100 g and 200 g are moving along the same line and direction with velocities of 2 m s and 1 m s , respectively. They collide and after the collision, the first object moves at a velocity of 1 . 67 m s . Determine the velocity of the second object.

Given data, Mass of the first object, m 1 = 100 g = 0 . 1 kg Mass of the second object, m 2 = 200 kg = 0 . 2 kg The velocity of the first object before the collision, v 1 = 2 m s The velocity of the second object before the collision, v 2 = 1 m s The velocity of the first object after the collision, v 3 = 1 . 67 m s The velocity of the second object after the collision, v 4 = ? We know that the momentum of an object measures the quantity of motion possessed by an object and it is calculated as the product of its mass and velocity. Also, in the given situation the velocity of second object can be determined by applying law of conservation of momentum. According to the law of conservation of momentum, Total momentum before collision = Total momentum after collision. m 1 v 1 + m 2 v 2 = m 1 v 3 + m 2 v 4 ⇒ 2 × 0 . 1 + 1 × 0 . 2 = 1 . 67 × 0 . 1 + v 4 × 0 . 2 ⇒ v 4 = 1 . 165 m s Hence, the velocity of the second object becomes 1 . 165 m / s after the collision.

E M B I B E

SCIENCE TEXTBOOK FOR CLASS IX>Chapter 9 - Force and Laws of Motion>Questions

NCERT Solutions for Chapter: Force and Laws of Motion, Exercise 2: Questions

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NCERT Science Solutions for Exercise - NCERT Solutions for Chapter: Force and Laws of Motion, Exercise 2: Questions

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Questions from NCERT Solutions for Chapter: Force and Laws of Motion, Exercise 2: Questions with Hints & Solutions

MEDIUM

9th CBSE

IMPORTANT

SCIENCE TEXTBOOK FOR CLASS IX>Chapter 9 - Force and Laws of Motion>Questions>Q 1

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

MEDIUM

9th CBSE

IMPORTANT

SCIENCE TEXTBOOK FOR CLASS IX>Chapter 9 - Force and Laws of Motion>Questions>Q 2

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

MEDIUM

9th CBSE

IMPORTANT

SCIENCE TEXTBOOK FOR CLASS IX>Chapter 9 - Force and Laws of Motion>Questions>Q 3

From a rifle of mass $4 kg$ , a bullet of mass $50 g$ is fired with an initial velocity of $35 \frac{m}{s}$ . Calculate the initial recoil velocity of the rifle.

MEDIUM

9th CBSE

IMPORTANT

SCIENCE TEXTBOOK FOR CLASS IX>Chapter 9 - Force and Laws of Motion>Questions>Q 4

Two objects of masses $100 g$ and $200 g$ are moving along the same line and direction with velocities of $2 \frac{m}{s}$ and $1 \frac{m}{s}$ , respectively. They collide and after the collision, the first object moves at a velocity of $1.67 \frac{m}{s}$ . Determine the velocity of the second object.

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