Question

A completely inelastic collision occurs between two balls of wet putty that move directly toward each other along a vertical axis. ?

A completely inelastic collision occurs between two balls of wet putty that move directly toward each other along a vertical axis. Just before the collision, one ball, of mass 3.0 kg, is moving upward at 18 m/s and the other ball, of mass 1.9 kg, is moving downward at 13 m/s. How high do the combined two balls of putty rise above the collision point? (Neglect air drag.)

Answer 1

`3.64m` rounded to two decimal places.

Explanation:

In a completely inelastic collision the colliding objects stick together after the collision and move together as a single object.

In the given problem, lets assume that the balls of putty are initially moving along the `y` axis, upward direction being the positive `y` direction. And the collision occurs at the origin of the coordinate system.

Initial Momentum `=m_1.v_1+m_2.v_2`

`=3.0xx18+1.9xx(-13)=29.3kgms^-1`

If `v_f` is the final velocity of the two-ball system after the collision then, Final momentum `=(m_1+m_2).v_f=(3.0+1.9)xxv_f=4.9xxv_f`

As momentum is conserved in inelastic collisions, setting both equal and solving for final velocity

`4.9xxv_f=29.3`, gives us
`v_f=29.3/4.9=5.98ms^-1` rounded to two decimal places.

To calculate the maximum height `h` attained by the combined system of two balls of putty after the the collision, we use the expression for linear motion under gravity; `g=9.81ms^-2`.

`v^2-u^2=2gh`,

Inserting the known values and noting that gravity is acting against the direction of motion.

`0^2-(5.98)^2=2xx(-9.81)xxh`, solving for `h` we obtain
`h=(5.98)^2/19.62=1.35m` rounded to two decimal places.