A) 100 m/s upward
B) 600 m/s upward
C) 100 m/s downward
D) 300 m/s upward
Correct Answer: A
Solution :
Velocity of particle after 5 s v = u - gt \[v=100-10\times 5\] \[=100\,-50=50\,m/s\] (upwards) Conservation of linear momentum gives \[Mv={{m}_{1}}{{v}_{1}}+{{m}_{2}}{{v}_{2}}\] ....(i) Taking upward direction positive, the velocity \[{{v}_{1}}\] will be negative. \[\therefore {{v}_{1}}=-25\,m/s,\,v=50\,m/s\] Also M = 1 kg, m1 = 400 g = 0.4 kg and \[{{m}_{2}}=(M-{{m}_{1}})=1-0.4=0.6\,kg\] Thus, Eq. (i) becomes, \[1\times 50=0.4\times (-25)+0.6\,{{v}_{2}}\] \[or50=-10+0.6\,{{v}_{2}}\] or \[0.6\,{{v}_{2}}=60\] \[or{{v}_{2}}=\frac{60}{0.6}=100\,m/s\] As \[{{v}_{2}}\] is positive, therefore the other part will move upwards with a velocity 100 m/s.
You need to login to perform this action.
You will be redirected in
3 sec
