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?

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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?

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Answer 1 · 2018-03-14T02:24:05

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

$v^{2} - u^{2} = 2 g h$ $v^2-u^2=2gh$

Inserting given values we get maximum speed of ball $(b)$ $(b)$ 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.

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.
$|Δp(a)|/|Δp(b)|=M/(2M)=1/2$

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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?

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