A ball of mass 0.01 kg is moving with a velocity of 50 m/s. On applying a
constant force for 2 seconds on the ball, it acquires a velocity of 70 m/s.
●Calculate the initial and final momentum of the ball.
●Also, calculate the rate of change of momentum and the acceleration of the
ball.
●What would be the magnitude of the force applied?
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Answers
Answer:
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Answer:
Answer
Given -
\implies⟹ m = 0.01 kg
\implies⟹ u = 50 m/s
\implies⟹ t = 2 sec
\implies⟹ v = 70 m/s
where
\longrightarrow⟶ v is final velocity
\longrightarrow⟶ u is initial velocity
\longrightarrow⟶ t is time
\longrightarrow⟶ m is mass of ball.
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To find -
1. p_1p
1
2. p_2p
2
3. p_2 - p_1p
2
−p
1
4. a
5. f
where
\longrightarrow⟶ p_1p
1
is initial momentum.
\longrightarrow⟶ p_2p
2
is \longrightarrow⟶ final momentum.
\longrightarrow⟶ a is acceleration.
\longrightarrow⟶ f is magnitude of force.
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Formula used -
\boxed{p = mv}
p=mv
\boxed{f = ma}
f=ma
\boxed{v = u + at}
v=u+at
where
\longrightarrow⟶ p is momentum.
\longrightarrow⟶ m is mass.
\longrightarrow⟶ v is velocity.
\longrightarrow⟶ u is initial velocity.
\longrightarrow⟶ a is acceleration.
\longrightarrow⟶ t is time.
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Solution -
1.
\longrightarrow⟶ m = 0.01 kg
\longrightarrow⟶ u = 50 m/s
\boxed{p_1 = mu}
p
1
=mu
\longrightarrow⟶ p_1p
1
= 0.01 × 50
Initial momentum = 0.5 kg m/s
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2.
\longrightarrow⟶ m = 0.01 kg
\longrightarrow⟶ v = 70 m/s
\boxed{p_2 = mv }
p
2
=mv
\longrightarrow⟶ p_2p
2
= 0.01 × 70
Final momentum = 0.7 kg m/s
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3.
\longrightarrow⟶ p_2 = 0.7 kg m/sp
2
=0.7kgm/s
\longrightarrow⟶ p_1 = 0.5 kg m/sp
1
=0.5kgm/s
p_2 - p_1p
2
−p
1
= 0.7 - 0.5 = 0.2 kg m/s
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4.
\longrightarrow⟶ v = 70 m/s
\longrightarrow⟶ u = 50 m/s
\longrightarrow⟶ t = 2sec
\boxed{v = u + at}
v=u+at
\longrightarrow⟶ 70 = 50 + 2a
\longrightarrow⟶ 20 = 2a
\longrightarrow⟶ a = 10 m/s²
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5.
\longrightarrow⟶ a = 10 m/s²
\longrightarrow⟶ m = 0.01 kg
\boxed{Force = ma}
Force=ma
\longrightarrow⟶ F = 10 × 0.01
\longrightarrow⟶ F = 0.1 N