What Velocity Does the Boy Gain After Throwing the Jug?

[GodAllen]

Homework Statement

A boy on roller blades throws a jug of water away from himself, giving it a speed of 14m/s. The boy's mass is 38kg and the mass of he jug and the water is 7.9kg. What is the velocity gained by the lad?

Homework Equations

Not sure.

The Attempt at a Solution

Not sure.

 

[dazedy]

This is a simple conservation of momentum problem, so assume that at the beginning, total momentum is equal to 0
Afterwards, the total momentum must stay the same.
Thus:
Momentum(before) = Momentum(boy) + Momentum(jug)
This should be enough to get you started

 

[mido808]

so you apply the formula P(before)=P(after). (And by the way the momentum before is 0 as the intial velocities of the boy and the jug are 0 as P(before)= m(boy)xv(boyintial)+m(jug)xv(jug int.))Hope u understood until now.
p(after)=m(boy)xv(boy final)+m(jug)xv(jug)
P(after)=38kg(v)+7.9kg(14m/s)
P(after)=38v+110.6
P(before)=p(after)
0=38v+110.6
38v=-110.6
V=-2.9
So velocity of the boy is 2.9 and its direction is in the opposite direction that he threw the jug in.
Thanks for reading in my answer and i hope u understand and everbody as well since i am a new user.

 

[ashishsinghal]

so you apply the formula P(before)=P(after). (And by the way the momentum before is 0 as the intial velocities of the boy and the jug are 0 as P(before)= m(boy)xv(boyintial)+m(jug)xv(jug int.))Hope u understood until now.
p(after)=m(boy)xv(boy final)+m(jug)xv(jug)
P(after)=38kg(v)+7.9kg(14m/s)
P(after)=38v+110.6
P(before)=p(after)
0=38v+110.6
38v=-110.6
V=-2.9
So velocity of the boy is 2.9 and its direction is in the opposite direction that he threw the jug in.
Thanks for reading in my answer and i hope u understand and everbody as well since i am a new user.

you are not supposed to solve the whole problem. this is against pf rules
https://www.physicsforums.com/showthread.php?t=414380

 

[rwisz]

As a scientist, the first step in solving this problem would be to identify the relevant equations and principles that can be applied. In this case, we can use the principle of conservation of momentum, which states that the total momentum of a system remains constant unless acted upon by an external force.

We can also use the equation for momentum, which is mass multiplied by velocity (p=mv).

Based on the given information, we can set up the equation:

Total initial momentum = Total final momentum

(mass of boy x velocity of boy) + (mass of jug and water x velocity of jug and water) = (mass of boy and jug and water x final velocity)

Solving for the final velocity, we get:

(38kg x 0m/s) + (7.9kg x 14m/s) = (38kg + 7.9kg) x final velocity

532.2kgm/s = 45.9kg x final velocity

Dividing both sides by 45.9kg, we get:

final velocity = 532.2kgm/s / 45.9kg = 11.6m/s

Therefore, the boy will gain a velocity of 11.6m/s in the direction opposite to the direction in which he threw the jug of water. This is due to the conservation of momentum, where the momentum of the boy and the jug and water combined is equal to the momentum of the boy after the throw.