A pan of mass `m = 1.5 kg` and a block of mass `M = 3 kg` are connected to each other by a light inextensible string, passing over a light pulley as shown in Fig. Initially, the block is resting on a horizontal floor. A ball of mass `m_0 = 0.5 kg` collides with the pan at a speed `v_0= 20 m//s`. Consider this instant of collision as `t = 0`. Assume collision to be perfectly inelastic. Now, Fig. answer the following questions based on the above information.

Find the velocity of pan + ball system at `t = 2.6 s`. Assume that the block comes to rest instantaneously after striking the floor.

Options

A. `4m//s` downward
B. `4m//s` upward
C. `0.6 m//s` upward
D. `0.4 m//s` downward

Answer

D

Explanation

Due to collision an impulsive tension occurs in the string.
For ball `+` pan system `-J=(m+m_(0)) v-m_(0)v_(0)`

For block `J=Mvimpliesv=2m//s`
As the string is taut and the mass of the block is greater than the mass of ball `+`pan, so the block and ball `+` pan are under deceleration given by
`a=(3g-2g)/5=2m//s^(2)`
So, block will again come to ground after time `t`, given by
`0=2t-1/2xx2t^(2)impliest=2s`
At this instant, the pan `+` ball system is moving with `v=2m//s` and the stirng gets slacked, so the system is moving under gravity.
To determine the speed of the system at `t=2.6 s`, first find the time after which string is again jerked. that will happen when pan `+` ball system crosses its `t=2s` instant during its downward journey, i.e., after its time of flight of motion under gravity.
`T=(2xx2)/10=0.4s`
i.e., string will again jerk at `t=2A`
For `tgt2.4s`, the system is again under a retardation of `2 m//s^(2)`. Before this, find the velocity of various component of system just after jerk.
We have
`-J'=2v'-2xx2` and `J=3v'`
`5v=4impliesv=4/5=0.8m//s` (downward)
The maximum height attained by theh block after the second jerk is given by `0=(0.8)^(2)-2xx2HimpliesH=0.16m`