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?​

Answer 1

Answer :-

• Initial momentum is 0.5kg m/s
• Final momentum is 0.7kg m/s
• Rate charge of momentum 0.1kg m/s²
• Acceleration is 10 m/s²
• Magnitude of force is 0.1 N

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 sec

_________________________________

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(70 - 50)/2

= 0.01(20)/2

= 0.01(10)

= 0.1 kg m/s²

Acceleration :-

According to first 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 = Rate of change of momentum

--» Force = 0.1 kg/m²

--»Force = 0.1 N