Question

Two objects of masses 1kg and 2kg moving with a velocities 2 m/s and 4m/s, respectively. They collide and after collision first object moves at a velocity 3m/s, then velocity of second object is *

Answer 1

Here,

$m_{1} = 1 \: kg \\ m_{2} = 2 \: kg$

and,

$u_{1} = 2\: m/s \\ u_{2} = 4\: m/s \\ v_{1} = 3\: m/s$

Since, No External force is being applied to the system, Therefore, the momentum of the system is conserved.

Applying Conservation of Momentum, We get

$P_{i} = P_{f}$

$= > m_{1}u_{1} + m_{2}u_{2} = m_{1}v_{1} + m_{2}v_{2}$

$= > (1 \times 2) + (2 \times 4) = (1 \times 3) + (2 \times v_{2})$

$= > 2 + 8 = 3 + 2v_{2} \\ = >v_{2} = (10 - 3) \div 2 = \frac{7}{3} m/s$

Hence after collision, the velocity of second object is 7/3 m/s.

Hope this helps.

Answer 2

Answer:

Two objects of masses 1kg and 2kg moving with velocities 2 m/s and 4m/s, respectively. They collide and the first object moves at a velocity of 3m/s, then the velocity of the second object is $3.5m/s$

Explanation:

• Momentum is conserved in every collision, that is after and before momentum should be the same

• When an object with masses $m_{1}$ with velocity $u_{1}$ and another object with mass $m_{2}$ and velocity $u_{2}$ collide with each other, after the collision the first object moves with velocity $v_{1}$ and the second one moves with velocity $v_{2}$,

• momentum before the collision is

        $m_{1} u_{1} +m_{2} u_{2}$

• momentum after collision is

        $m_{1} v_{1} +m_{2} v_{2}$

• then according to the momentum conservation

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

From the question, we have

$m_{1} =1kg$

$m_{2} =2kg$

$u_{1} =2m/s$

$u_{2} =4m/s$

$v_{1} =3m/s$

put the values to get the velocity of the second object after the collision

$1*2+2*4=1*3+2*v_{1}$

⇒

$7=2v_{1} \\v_{1}=7/2=3.5m/s$