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# Two objects of masses 100 g and 200 g are moving along the same line and direction with velocities of 2 m s^(−1) and 1 m s^(−1), respectively. - CBSE Class 9 - Science

ConceptConservation of Momentum

#### Question

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

#### Solution

Mass of one of the objects, m1 = 100 g = 0.1 kg

Mass of the other object, m2 = 200 g = 0.2 kg

Velocity of m1 before collision, v1 = 2 m/s

Velocity of m2 before collision, v2 = 1 m/s

Velocity of mafter collision, v3 = 1.67 m/s

Velocity of mafter collision = v4

According to the law of conservation of momentum:-

Total momentum before collision = Total momentum after collision

∴ m1v1 + m2v2 = m1v3 + m2v4

(0.1)2 + (0.2)1 = (0.1)1.67 + (0.2)v4

0.4 = 0.167 + 0.2v4

∴ v4 = 1.165 m/s

Hence, the velocity of the second object becomes 1.165 m/s after the collision.

Is there an error in this question or solution?

#### APPEARS IN

NCERT Solution for Science Textbook for Class 9 (2018 to Current)
Chapter 9: Force and Laws of Motion
Q: 4 | Page no. 127

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Solution Two objects of masses 100 g and 200 g are moving along the same line and direction with velocities of 2 m s^(−1) and 1 m s^(−1), respectively. Concept: Conservation of Momentum.
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