Question

A batsman hits back a ball straight in the direction of the bowler without changing its initial speed of 12m/s. If the mass of the ball is 0.15kg . Determine the impulse imparted to the ball

Answer 1

Answer: 3.6 kg m/s

Step-by-step explanation:

Impulse = Change in Momentum = mv - mu

m(v - u)

Since the batsman hits it straight towards the bowler, the velocity will be in negative.

Taking u as - 12 m/s and v as 12 m/s, (Depends upon which direction tu take as positive)

Impulse = 0.15 ( 12 - [-12])

= 0.15 × 24

= 3.6 kg m/s

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Answer 2

$\sf \fcolorbox{red}{pink}{ \huge{Solution :)}}$

Given ,

• Initial speed of ball (u) = 12 m/s
• Final speed of ball (v) = - 12 m/s
• Mass of ball (M) = 0.15 kg

As we know that ,

$\large \sf \fbox{IMPULSE = CHANGE \: IN \: MOMENTUM </p><p>}$

It can be written as ,

$\large \sf \fbox{IMPULSE = Mu - Mv</p><p>}$

Substitute the known values , we get

Impulse = 0.15(12) - 0.15(-12)

Impulse = 0.15(12 + 12)

Impulse = 0.15(24)

Impulse = 3.6 N•s or Kg•m/s

Hence , the 3.6 N•s impulse imparted to the ball

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