Question

4> 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?

Answer 1

Answer:

answer given below

Explanation:

p=mv

{f = ma}f=ma

{v = u +

⟶ p is momentum.

⟶ m is mass.

⟶ v is velocity.

⟶ u is initial velocity.

⟶ a is acceleration.

⟶ t is time.

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Solution -

1.

⟶ m = 0.01 kg

⟶ u = 50 m/s

{p_1 = mu}p1=mu

⟶ p_1p1 = 0.01 × 50

Initial momentum = 0.5 kg m/s

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2.

⟶ m = 0.01 kg

⟶ v = 70 m/s

{p_2 = mv }p2=mv

⟶ p_2p2 = 0.01 × 70

Final momentum = 0.7 kg m/s

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3.

⟶ p_2 = 0.7 kg m/sp2=0.7kgm/s

⟶ p_1 = 0.5 kg m/sp1=0.5kgm/s

p_2 - p_1p2−p1 = 0.7 - 0.5 = 0.2 kg m/s

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4.

⟶ v = 70 m/s

\longrightarrow⟶ u = 50

⟶ t = 2sec

{v = u + at}v=u+at

⟶ 70 = 50 + 2a

⟶ 20 = 2a

⟶ a = 10 m/s²

━━━━━━━━━━━━━

5.

⟶ a = 10 m/s²

⟶ m = 0.01 kg

{Force = ma}Force=ma

⟶ F = 10 × 0.01

⟶ F = 0.1 N

Answer 2

Answer:

Initial momentum is $0.5 kg m/s$

Final momentum is $0.7 kg m/s$

The rate of change of momentum is $0.2 kg m/s$

Acceleration is $10 m/s^{2}$

Force applied is $0.1 N$

Explanation:

1. Momentum is product of mass and velocity

$p=mu=0.01 * 50=0.5 kg m/s$

Initial momentum is $0.5 kg m/s$

2. Final velocity is $70m/s$

Hence, final momentum is $mv= 0.01 * 70= 0.7 kg m/s$

3. Change in momentum is difference in final and initial momentum.

$= 0.7 - 0.5 = 0.2 kg m/s$

4. Acceleration is calculated using $v=u+at$

$70 = 50 + 2a$

$a=10m/s^{2}$

5. Force is $F=ma$

$F=10 * 0.01$

$F= 0.1 N$

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