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.

Mark the correct statement(s) for this situation

Options

A. After the collision, the pan `+` ball system moves downwards with decreasing speed.
B. After the collision, the block is moving upwards with the same speed with -which the ball `+` pan system is moving downwards
C. The block will jerk for a number of times during its motion.
D. All of these

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`