Question

A 5kg ball rolls on a smooth surface with a velocity 10 metre per second collides with another ball b of mass 10 kg moving opposite to balll A with a velocity 4 metre per second as shown in the following figure. The coefficient restitution age 0.6 between both the balls .calculate the velocities of balls immediately after impact

Answer 1

Explanation:

e=velocity of seperation÷velocity of approach

next we need to total their MOI

M1V1 + M2V2=M1U1 + M2U2

Answer 2

The velocities of balls after impact is -4.94 m/s and 3.46 m/s.

Explanation:

Given that,

Mass of ball = 5 kg

Velocity of ball = 10 m/s

Mass of another ball = 10 kg

Velocity of another ball = 4 m/s

Coefficient restitution = 0.6

We need to calculate the  velocity

Using equation of coefficient restitution

$v_{2}-v_{1}=e(u_{1}+u_{2})$

Put the value into the formula

$v_{2}-v_{1}=0.6(10+4)$

$v_{2}-v_{1}=8.4$....(I)

We need to calculate the  velocity

Using conservation of momentum

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

Put the value into the formula

$5\times10-10\times4=5v_{1}+10v_{2}$

$v_{1}+2v_{2}=2$....(II)

From eqyation (I) and (II)

$3v_{2}=10.4$

$v_{2}=\dfrac{10.4}{3}$

$v_{2}=3.46\ m/s$

Put the value of v₂ in equation (I)

$3.46-v_{1}=8.4$

$v_{1}=-4.94\ m/s$

Hence, The velocities of balls after impact is -4.94 m/s and 3.46 m/s.

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