Question

A body of mass 0.25kg moving with velocity 12m/s is stopped by applying a force of 0.6N.calculate the time taken to stop the body.also calculate the impulse of this force

Answer 1

1st method

Mass of the body (m) = 0.25 kg 
Initial Velocity (u) = 12 m/s

Force applied to stop that body (F) = -0.6 N (- sign shows that force is applied in the opposite direction of motion of that body)

Now, we know that,
Since - a = F/ m
            a = - 0.6/ .25
            a =-12/5

According to third equation of motion
v^2 = u^2 + 2as
(o) = (12)^2 + 2 (-12/5) s
(0) = 144 - 2(12/5) s
2(12/5) s = 144

s = 144 x 5/ 24
s = 30 m

 Now according to second equation of motion-
s = ut + 1/2 at^2

30 = 12t + 1/2(-12/5) t^2 

30 = 12t - 6t^2/5

6t^2/5 - 12t + 30 =0

6(t^2/5 - 2t + 5) = 0

t^2 - 10t + 25 = 0

t^2 - 5t - 5t +25 =0

t(t-5) -5(t-5)

so both side come t-5 = 0
 t = 5 sec

2nd method

Explanation: m = .25kg, u = 12m/s, v = 0, F = -0.6N
a = F/m = 0.6/0.25 = -24m/s2
t = (v-u)/a = (0-12)/(-24) = 5s.


and impulse

I = F x s = -0.6 x 5 = -3 Ns


i hope it will help you
regards