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
Assuming origin to be at the point of projection and downwards being the direction of
#v^2-u^2=2gh#
Inserting given values we get maximum speed of ball
#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
#|Δp(b)|=2Msqrt(2gH)# ........(2)
Given initial speed of ball
The vertical motion of ball
#vecv_a=sqrt(2gH)/2hati-sqrt(2gH)\ hatj#
Change in velocity of ball
#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.
- Since movements along
#xandy# are orthogonal these can be treated independently. - 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# .) - 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# - Both balls are falling freely under gravity in this direction.
#|Δp(a)|/|Δp(b)|=M/(2M)=1/2#