Question

a ball of mass 10g is moving with a speed of 30m/s.On applying a constant force on ball for 2.0s, acquire a velocity of 70 m/s

1)initial momentum
2)final momentum
3)the rate of change in momentum
4) acceleration
5)the magnitude of force

Answer 1

HELLO THERE!

Given data:

Mass of the ball (m) = 10g = 0.01kg

Initial velocity (u) = 30 m/s

(1) So, Initial momentum (p1) = mu

= 0.01kg x 30 m/s

= 0.3 kg m/s

Final velocity (v) = 70 m/s

(2) So, final momentum (p2) = mv

= 0.01 kg x 70 m/s

= 0.7 kg m/s

Time for which the constant force is applied = 2 seconds.

(3) Rate of change of momentum =

$\frac{p_{2} - p_{1}}{t}$

$= \frac{0.7 - 0.3}{2} \\\\= 0.2 kgm^{2}s^{-2}$

(5) The magnitude of force:

Impulse = Force x Time interval (as we know).

Further, Impulse = Change in Momentum.

So, Force x Time interval = Change in momentum

=> Force = (0.7 - 0.3)/2

= 0.2 N

(This answer satisfies the Newton's Second Law of Motion, which states that the force applied on a body is directly proportional to its change in momentum).

(4) Acceleration of the body when the force is applied:

From Newton's Second Law of Motion,

F = ma

So, a = 0.2/0.01

= 20 m/s²

THESE ARE YOUR ANSWERS..

Thanks!