Question

A force of 24 N acts on a body of mass 6 kg for 4 s. Assuming the body to be initially at rest. Then (A) When the force stops acting its velocity is 16 m/s (B) The distance covered in 10 s after the force acting is 128 m (C) The distance covered in 5 s after the force stops acting is 96 m (D) Its momentum after 10 s is 96 kg ms The correct option(s) is/are i)Only B ii)Only A iii) Both (B) and (D) iv) (A), (B) and (D)​

Answer 1

Answer:

A,B<D

Explanation:

F=ma --->24=6.a then a=4 m/s^-2

i.v=u+at since start from rest u=0

therefore after 4 s(when force stops acting)

v=4.4=16 m/s(Option A)

ii.distance covered in 10 s.

for the first 4 s,distance travelled by the action of  force .

v^2-u^2= 2as on putting values as v=16 m/s,u=0,a=4 m/s^2 then

S =32 m,

as force stops acting after 4 s,body moves with constant velocity(NLM-I) then acceleration=0.

then distance travelled is

S'=vt=16.6

=96 m

therefore,total distance covered is S=S(4 s)+S'(6 s)

=32+96=128 m(option B)

iii.as force stops acting,according to NLM-I, Fnet=0 then P=Constant

Momentum is m(v-u)=6(16-0)=96 kg m/s (Option D)

this momentum is same for both  4 s and  10 s.