what impluse is needed to stop a ball of mass 200g travelling a velocity 110 kmph
Answers
Answer:
Impulse is defined as the change in momentum that arises due to some applied force under a particular time interval.
⇒ ΔP = FΔT
⇒ ΔP = m.a.Δt
⇒ ΔP = m.Δv
According to the question, it is given that a ball of mass 200g (0.2 kg) is travelling with a velocity 110 kmph. Let us first convert the velocity to m/s.
⇒ 110 kmph × (5/18) = 550/18
⇒ Velocity = 30.56 m/s
Now we know that, ball is being stopped. Hence the final velocity would be 0 m/s. Applying it in the formula we get:
⇒ ΔP = mΔv
⇒ ΔP = 0.2 ( 0 m/s ) - 0.2 ( 30.56 )
⇒ ΔP = 0 - 6.112
⇒ ΔP = -6.112 Kg.m/s
Hence the impulse required to stop the ball is 6.112 Kg.m/s in the opposite direction.
✠ Mass of ball = 200 grams.
✠ Velocity of ball = 110 km/h
✠ Impulse needed to stop the ball.
✠ Impulse needed to stop the ball = 6.112 kg.m/s
✠ ∆P (To find impulse)
✠ Impulse = m∆v
~ As it's already given that the mass of the ball is 200 grams. Means let's convert gram into kilogram firstly.
✨ Converting g into kg
➙ 1 kg = 1000 grams
➙ 1 gram = 1/1000 kg
➙ 200 grams = 200/1000
➙ 0.2 kilograms
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~ It is also given that it is travelling with a velocity of 110 km/h so let's convert it into m/s.
✨ Converting km/h into m/s
➙ 110 km/h = (5/18) = 550/18
➙ Velocity = 30.56 m/s
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~ It is also given that the ball is about to stop henceforth, it's final velocity be 0 m/s by itself..!
✨ Let's apply the formula..!
➙ ∆P = m∆v
➙ ∆P = 0.2(0)-0.2(30.56)
➙ ∆P = 0-6.112
➙ ∆P = -6.112
But we haven't to take impulse as negative so let's take it as positive..!
➙ ∆P = 6.112 kg.m/s
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