Momentum Theory question Gr.12

[elmosworld403]

1. A student on a skateboard, with a combined mass of 78.2 kg, is moving east at 1.60 m/s. As he goes by, the student skilfully scoops his 6.4-kg backpack from the bench where he had left it. What will be the velocity of the student immediately after the pickup?






Me solving
momentum of student = momentum of student+bag
MV=MV
78.2kg(1.60 m/s E)= 84.6kg( Velocity)

125.12=84.6(Velocity)
Velocity=1.478959811
Velocity=1.5 m/s (E)


QUESTION- What I don't get is the theory behind this. How does the momentum of the student equal the momentum of student and bag? Please explain how this works.

 

[Bradyns]

1. A student on a skateboard, with a combined mass of 78.2 kg, is moving east at 1.60 m/s. As he goes by, the student skilfully scoops his 6.4-kg backpack from the bench where he had left it. What will be the velocity of the student immediately after the pickup?
Me solving
momentum of student = momentum of student+bag
MV=MV
78.2kg(1.60 m/s E)= 84.6kg( Velocity)

125.12=84.6(Velocity)
Velocity=1.478959811
Velocity=1.5 m/s (E)QUESTION- What I don't get is the theory behind this. How does the momentum of the student equal the momentum of student and bag? Please explain how this works.

(m_s×v_s)+(m_b×v_b) = (m_s+m_b)v^2

Where:
m_s = Mass of student
v_s = Velocity of student
m_b = Mass of bag
v_b = Velocity of bag

In writing;
The sum of the initial momentum (p=mv) on both objects is equal to the final momentum, where the masses are combined.

If you have objects moving on angles, you can find the (x,y) components of the velocity vectors to determine what occurs. x=[(v)cos(θ)], y=[(v)sin(θ)]