A 8 kg ball moving with a velocity of 4 m/s collides with a 2 kg ball moving with a velocity of 8 m/s in the opposite direction. If the collision is perfectly elastic, the velocities of balls after the collision will be ____

Question

A 8 kg ball moving with a velocity of  4 m/s collides with a 2 kg ball moving with a velocity of  8 m/s in the opposite direction. If the collision is perfectly elastic, the velocities of balls after the collision will be ____

Infinity Learn · Accepted Answer

If the collision is perfectly elastic, the velocities of balls after the collision will be

- \frac{4}{5} m / s

and

\frac{56}{5} m / s .

According to the principle of conservation of momentum, the total linear momentum of an
isolated system, i.e., a system for which the net external force is zero, is constant.
Given
Mass of the first ball,

M_{1} = 8 kg .

Velocity of the first ball,

V_{1} = 4 m / s .

Mass of the second ball,

M_{2} = 2 kg .

Velocity of the second ball,

V_{2} = 8 m / s

Total momentum

= M_{1} V_{1} + M_{2} V_{2}

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

8 \times 4 - 2 \times 8 = 8 v_{1} + 2 v_{2} \dots (i)

As collision is elastic, so we have

\frac{1}{2} m_{1} {u_{1}}^{2} + \frac{1}{2} m_{2} {u_{2}}^{2}

=

\frac{1}{2} m_{1} {v_{1}}^{2} + \frac{1}{2} m_{2} {v_{2^{}}}^{2}

\frac{1}{2} \times 8 \times 4^{2} + \frac{1}{2} \times 2 \times 8^{2}

=

\frac{1}{2} \times 8 {v_{1}}^{2} + \frac{1}{2} \times 2 {v_{2^{}}}^{2}

—-(ii)
Solving equations (i) and (ii),
we get

v_{1} = - \frac{4}{5} m / s

and,

v_{2} = \frac{56}{5} m / s .

A $8 kg$ ball moving with a velocity of $4 m / s$ collides with a $2 kg$ ball moving with a velocity of $8 m / s$ in the opposite direction. If the collision is perfectly elastic, the velocities of balls after the collision will be ____

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