Question

The velocity of a body of mass 10 kg increases from 4 m/s to 8 m/s. When a force act on it for 2 seconds, find : (a) What is the momentum before the force act? (b) What is the momentum after the force act? (c) What is the momentum gain in momentum per sec? (d) What is the value of force?

Answer 1

(a) momentum before the force
=mass × initial velocity
=10×4
=40kg m/s
(b) momentum after the force
=10×8
=80kg m/s
(c)rate of change of momentum
=80-40/2
=20 kg m/s²
(d) force
= ma
=m(v-u)/t
=10(4)/2
=20N

Answer 2

Given ;
m = 10 kg
v1 = 4m/s
v2 = 8m/s
t = 2 sec

(a) Momentum before the act of force
= mv1
= 10×4
= 40 kg.m/s

(b) Momentum after the act of force
= mv2
= 10×8
= 80 kg.m/s

(c) Momentum gain per sec
= Final Momentum-Initial Momentum/Time
= (mv2-mv1)/t
= (80-40)/2
= 20 kg.m/s^2

(d) Force
= m×a
= m (v2-v1)/t
= 10×(8-4)/2
= 20 N