Question

Mass m = 0.1 kg moves to the right with speed v = 0.45 m/s and collides with an equal mass initially at rest. After this inelastic collision the system retains a fraction = 0.86 of its original kinetic energy. If the masses remain in contact for 0.01 secs while colliding, what is the average force in N between the masses during the collision?

Answer 1

`F = 4.16 N`

Step-by-step explanation:

Let the coefficient of restitution is "e"

so we have

`e = (v_2 - v_1)/(u_1 - u_2)`

here we have

`v_2 - v_1 = eu`

now we have by momentum conservation

`mu = mv_1 + mv_2`

`v_1 + v_2 = u`

now we have

`v_2 = (u)/(2)(1 + e)`

`v_1 = (u)/(2)(1 - e)`

now we have the ratio of final kinetic energy and initial kinetic energy given as 0.86

so we have

`0.86 = (0.5mv_1^2 + 0.5mv_2^2)/(0.5 mu^2)`

`0.86 u^2 = v_1^2 + v_2^2`

`0.86 u^2 = (u^2)/(4)(1 + e)^2 + (u^2)/(4)(1 - e)^2`

`3.44 = 2 + 2e^2`

`e = 0.85`

so we have

`v_1 = (u)/(2)(1 - 0.85) = 0.076u`

now change in velocity of first ball is given as

`\Delta v = u - 0.076 u`

`\Delta v = 0.924u`

now acceleration of the ball is given as

`a = (\Delta v)/(\Delta t)`

`a = (0.924* 0.45)/(0.01)`

`a = 41.6 m/s^2`

so the force on the ball during collision is

`F = ma`

`F = 0.1 * 41.6`

`F = 4.16 N`