From a height H, I shoot horizontally a ball(a) (mass M), while I drop another ball (b) with mass (2M). The initial speed of ball (a) is half of the max speed of ball(b). What is the ratio |Δp(a)|/|Δp(b)| over the whole time of motion?

1 Answer
A08
Mar 14, 2018

Assuming origin to be at the point of projection and downwards being the direction of #-hatj#. For ball #(b)# ignoring air friction kinematic expression is

#v^2-u^2=2gh#

Inserting given values we get maximum speed of ball #(b)# as it touches ground

#v_b^2-0^2=2(-g hatj)cdot(-H\ hatj)#
#v_b^2=2gH#
#vecv_b=-sqrt(2gH)\ hatj# .......(1)
selected #-ve# root in line with the defined direction.

As the initial speed is #=0#, change in magnitude of momentum

#|Δp(b)|=2Msqrt(2gH)# ........(2)

Given initial speed of ball #(a)# #=sqrt(2gH)/2#
#=>vecu_a=sqrt(2gH)/2hati# .......(3)

The vertical motion of ball #(a)# will have the same velocity as in (1). As we have ignored air friction, there is no change in its velocity along the #x#-axis. Final velocity of ball #(a)# is

#vecv_a=sqrt(2gH)/2hati-sqrt(2gH)\ hatj#

Change in velocity of ball #(a)#

#vecv_a-vecu_a=(sqrt(2gH)/2hati-sqrt(2gH)\ hatj)-sqrt(2gH)/2hati#
#vecv_a-vecu_a=-sqrt(2gH)\ hatj#
#:.|Δp(a)|=Msqrt(2gH)# ............(4)

From (2) and (4) desired ratio is

#|Δp(a)|/|Δp(b)|=1/2#

.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.--.

For this question we can arrive at the same result by reasoning.

  1. Since movements along #xandy# are orthogonal these can be treated independently.
  2. Ball #(a)#, ignoring air friction, there is no change of velocity in #x#-direction. Only change in velocity is in the #y#-direction. Which is independent of mass. Initial velocity #=0#.
    Therefore #|Deltav(a)|=sqrt((Deltav_x)^2+(Deltav_y)^2)=v_y#, (second term in the magnitude expression #=0#.)
  3. Ball #(b)#, there is no velocity in #x#- direction. Only change in velocity is in the #y#-direction. Which is independent of mass. Therefore it is same as for ball #(a)#. #|Deltav(b)|=v_y#
  4. Both balls are falling freely under gravity in this direction.
  5. #|Δp(a)|/|Δp(b)|=M/(2M)=1/2#
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