a skateboarder with mass 75 kg is riding their skateboard of mass 8 kg at 1.5 m/s. If they decide to jump off sp that their body has no horizontal movement in the air before landing, how fast must the skateboard be moving after they separate
Initial momentum=final momentum
(75+8)1.5=8*V
solve for V
To solve this problem, we can apply the principle of conservation of momentum. In the absence of any external forces, the total momentum before the separation will be equal to the total momentum after the separation.
The momentum of an object is calculated by multiplying its mass by its velocity. Let's denote the velocity of the skateboarder as Vskateboarder, the velocity of the skateboard as Vskateboard, and the final velocity (after separation) of the skateboard as Vfinal_skateboard.
The total momentum before separation is given by the sum of the momentum of the skateboarder and the momentum of the skateboard:
Momentum before = (mass of skateboarder) * (velocity of skateboarder) + (mass of skateboard) * (velocity of skateboard)
Momentum before = (75 kg) * (1.5 m/s) + (8 kg) * (0 m/s) (since the skateboard has no initial velocity)
Now we know that after separation, the skateboarder has no horizontal movement, implying that the skateboard gains all the horizontal momentum. Therefore, the momentum after separation will be equal to the momentum of the skateboard:
Momentum after = (mass of skateboard) * (velocity of skateboard) = (8 kg) * Vfinal_skateboard
Setting the two momenta equal to each other, we get:
Momentum before = Momentum after
(75 kg) * (1.5 m/s) + (8 kg) * (0 m/s) = (8 kg) * Vfinal_skateboard
Now we can solve for Vfinal_skateboard:
112.5 kg·m/s = 8 kg · Vfinal_skateboard
Vfinal_skateboard = 112.5 kg·m/s / 8 kg = 14 m/s
Therefore, after the skateboarder separates from the skateboard, the skateboard must be moving at a speed of 14 m/s in order for the skateboarder's body to have no horizontal movement in the air before landing.