How to Calculate Horizontal Jumping Speed on a Skateboard?

[slyskittle]

if someone was standing on a skateboard at rest and then jumped off, how would you calculate the person's horizontal jumping speed (if the skateboard rolled in the opposite direction). what information would i need? (the masses of the person and the skateboard and the speed which the skateboard went?)
thanks for any help!

 

[dst]

Conservation of momentum - (total mass of skateboarder + board) * velocity = velocity of skateboarder * his mass + velocity of board + its mass.

For instance, 65kg boarder, 5kg board, initially at 10m/s. Initial momentum = (65+5)kg*10m/s = 700kgm/s. Then assume the board stopped completely - 65kg boarder now takes on all the momentum, his speed will be 700kgm/s divided by 65kg - approx 10.76m/s.

On the other hand, unseatbelted dude and car - momentum = 60kg + 1000kg * 30m/s = 82400kgm/s. The car barely has to slow down to kill said dude.

 

[astrokreng]

To calculate the horizontal jumping speed of a person on a skateboard, we would need to use the conservation of momentum principle. This principle states that the total momentum of a system remains constant unless acted upon by an external force. In this case, the system consists of the person and the skateboard.

To begin, we would need to gather the necessary information, which includes the masses of the person and the skateboard, and the speed at which the skateboard went in the opposite direction. Let's say the mass of the person is represented by m1, the mass of the skateboard is represented by m2, and the speed of the skateboard in the opposite direction is represented by v2.

Next, we would use the formula for conservation of momentum, which is:

m1v1 + m2v2 = m1v1' + m2v2'

Where:
m1 = mass of the person
v1 = initial velocity of the person (at rest)
m2 = mass of the skateboard
v2 = initial velocity of the skateboard (opposite direction)
v1' = final velocity of the person (after the jump)
v2' = final velocity of the skateboard (after the jump)

Since the person is jumping off the skateboard, the final velocity of the skateboard would be zero (v2' = 0). Therefore, the equation becomes:

m1v1 + m2v2 = m1v1' + 0

Solving for v1', we get:

v1' = (m1v1 + m2v2)/m1

This would give us the final velocity of the person after the jump. To calculate the horizontal jumping speed, we would need to use the Pythagorean theorem, which states that the square of the hypotenuse (jumping speed) is equal to the sum of the squares of the other two sides (vertical and horizontal velocities). This can be represented by the equation:

v^2 = v1'^2 + v2'^2

Where:
v = horizontal jumping speed
v1' = final velocity of the person (calculated earlier)
v2' = final velocity of the skateboard (zero in this case)

Solving for v, we get:

v = √(v1'^2 + v2'^2)

Therefore, to calculate the horizontal jumping speed of the person on the skateboard, we would need the masses of the person and the skateboard, and the speed at