Question

A hokey ball of mass 200g hits on a hockey stick with a velocity 10m/s . calculate the change in momentum if the ball bounces back on the same path with same speed ?
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Answer 1

Answer:

mass of ball = 200g = 0.2 kg

initial velocity = 10m/s

final velocity = -5m/s

initial momentum = 0.2 × 10

= 2Ns

final momentum = 0.2kg × -5m/s

= -1Ns

change in momentum = final momentum - initial momentum

= -1Ns -2Ns = 3Ns

explanation:

Answer 2

If a hokey ball of mass 200g hits on a hockey stick with a velocity 10m/s, bounces back on the same path with same speed  then change in momentum is $4\hspace{1 mm}kgms^{-1}$

Step-by-step Explanation:

This question uses the concept of Linear momentum, which is derived from Newton's second law of motion.

Given:

• Mass of ball $m=200\hspace{1 mm} g= 0.2 \hspace{1 mm} kg$
• Initial velocity $v_1=-10 \hspace{1 mm} m/s$ (downward direction, so by sign convention: negative)
• Final velocity $v_2= 10 \hspace{ 1mm} m/s$ (upward direction)

To be Found:

Change in momentum= $m(v_2-v_1)$

Concepts used:

• Newton's second law of motion: Force is equal to rate of change of momentum
• Linear momentum is defined as scalar product of mass with the velocity (but it is a vector quantity) $p=m.v$

• Change in momentum= $Final \hspace{1 mm}p- Initial \hspace{1 mm}p = mv_2-mv_1=m(v_2-v_1)$  

• According to a sign convention, we can choose any direction to be positive. Here we can choose upward direction to be positive. So, the downward direction automatically becomes negative. This is needed as Momentum is a vector quantity.

We can observe here, Change in momentum = $m(v_2-v_1)= 0.2 * (10-(-10)) = 0.2*(10+10) = 0.2 * 20 = 4 \hspace{1 mm}kgms^{-1}$ (in upward direction)

So, if a hokey ball of mass 200g hits on a hockey stick with a velocity 10m/s, and bounces back on the same path with same speed  then change in momentum is $4\hspace{1 mm}kgms^{-1}$