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.

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Solution

given
$m_{1} = 0.01 k g$
$m_{2} = 0.02 k g$
$u_{1} = 2 m / s$
$u_{2} = 1 m / s$
$v_{1} = 1.67 m / s$
from law of conservation of energy
$m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2}$
$0.01 \times 2 + 0.02 \times 1 = 0.01 \times 1.67 + 0.02 v_{2}$
$v_{2} = 1.165 m / s$

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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 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.

givenm1=0.01kgm2=0.02kgu1=2m/su2=1m/sv1=1.67m/sfrom law of conservation of energym1u1+m2u2=m1v1+m2v20.01×2+0.02×1=0.01×1.67+0.02v2v2=1.165m/s

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.

given
$m_{1} = 0.01 k g$
$m_{2} = 0.02 k g$
$u_{1} = 2 m / s$
$u_{2} = 1 m / s$
$v_{1} = 1.67 m / s$
from law of conservation of energy
$m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2}$
$0.01 \times 2 + 0.02 \times 1 = 0.01 \times 1.67 + 0.02 v_{2}$
$v_{2} = 1.165 m / s$