a trolley of mass 200 kg moves with a uniform speed of 36 kmhr on a frictionless track. a child of mass 20 kg runs on d trolley from one end to the other (10 m away) with a a speed of 4m/s relative to the trolley in a direction opposite to its motion, and jumps out of the trolley. What is the final speed of the trolley ? how much has the trolley moved from the time the child begins to run ??
6anyone of you can answer it? why the boys mass not included in the initial momentum?
3sky girl ek concept galat hai tera...........
see as soon as the boy starts runnin the trolley will experience a recoil velocity ...............
not after he jumps out!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
why?? basically because of conservation of momentum!!!!!!!!!!!!!
6if it was to be discussed in my school class room my teacher would have made it round off to 10.4m/s and i got 10.44m/s which is also the same and hence no more discussion and fuss in the classs! ask any XItheee!
6well i got 10.44! and thats according to IITCOMING's eqauation(when include the boy's mass in the LHS)
6well if we include the boy's mass in the equation of initial momentum i got the answer to be 37.6 km/hr! what is the correct answer?
1i should be included ... main khud nahi janti usne kya likha hai ... [2]
how i did :
200X10 + 20X6 = 200 V
=> v =10.6 M/S
book me hai: 10.36 m/s ...
3BASICALLY U CONSERVE MOMENTUM ALONG HORIZONTAL DIRECTION!!!!!!!!!!!!!!!!!!!!!!!!
IVE GTG NOW ILL TRY TO SOLVE IT AFTER I CUM BAK!!!!!!![2]
1whom u ar asking ??
well iitim... ,, ans galat hi ara tumhare eqn se...
6well can anyone explain me why the boy's mass not included in the initial momentum of the system though he is on the trolley!
6what is your reference frame? are you getting it done from the ground reference of from the trolley's reference?
1arey woh toh pata hi hai!!
y recoil vel here ??
recoil toh kareg a afetr boy jumps out...
y recoil in boy's vel ??
6@skygal: did you get why the child's mass is excluded in the LHS?
3u are supposed to find the velocity of the child WRT the ground!!!!!!![1]
3220(10) = 20(10+Vr-4) + 200(10+Vr)
2200 = 120 + 20Vr +2000 + 200Vr ......
2200 - 2120 = 220Vr ....
80=220Vr ...
4/11 = Vr ....
0.363=Vr...
velocity of truck = 10+0.363....
10.363.......