Question

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?​i​

Answer 1

Answer:

Momentum= mass* velocity

acceleration=velocitu/time

force= mass*acceleration

Explanation:

1) initial momentum= 0.01×50=0.5

Final momentum= 0.01×70=0.7

2)rate of change of momentum= 0.7-0.5=0.2

acceleration=70 / 2= 35 m/s^2

3) force= 0.01× 35=0.35 newton

Answer 2

Answer :-

• Initial momentum is 0.5 kg m/s .

• Final momentum is 0.7 kg m/s .

• Rate change of momentum 0.1 kg m/s² .

• Acceleration is 10 m/s² .

• Magnitude of force is 0.1 Newton .

Explanation :-

We have :-

→ Mass of ball (m) = 0.01 kg

→ Initial velocity (u) = 50 m/s

→ Final velocity (v) = 70 m/s

→ Time (t) = 2 seconds

________________________________

Initial momentum :-

= m × u

= 0.01 × 50

= 0.5 kg m/s

Final momentum :-

= m × v

= 0.01 × 70

= 0.7 kg m/s

Rate of change of momentum :-

= (mv - mu)/t

= m(v - u)/t

= 0.01(70 - 50)/2

= 0.01(20)/2

= 0.01(10)

= 0.1 kg m/s²

Acceleration :-

According to 1st equation of motion .

⇒ v = u + at

⇒ 70 = 50 + a(2)

⇒ 70 - 50 = 2a

⇒ 20 = 2a

⇒ a = 20/2

⇒ a = 10 m/s²

Magnitude of force :-

According to Newton's 2nd law of motion we know that :-

Force = Mass × Acceleration

⇒ F = 0.01 × 10

⇒ F = 0.1 N