A 55 kg person runs with a horizontal velocity of 8.0 m/s and jumps onto a 7 kg skateboard
which is initially at rest. Assuming negligible friction, what will be the final velocity of the
skateboard and the person?
To determine the final velocity of the person and the skateboard, we can use the principle of conservation of momentum. According to this principle, the total momentum before the interaction equals the total momentum after the interaction, assuming no external forces act on the system.
The momentum of an object is calculated by multiplying its mass by its velocity:
momentum = mass * velocity
Considering that the person and the skateboard are initially separate entities, the total momentum before the interaction is the sum of their individual momenta:
initial momentum = (mass of person) * (initial velocity of person) + (mass of skateboard) * (initial velocity of skateboard)
To find the final velocity, we divide the initial momentum by the total mass of the person and the skateboard:
final velocity = (initial momentum) / (total mass of person and skateboard)
Now let's calculate the values:
Mass of person (m1) = 55 kg
Initial velocity of person (v1) = 8.0 m/s
Mass of skateboard (m2) = 7 kg
Initial velocity of skateboard (v2) = 0 m/s (at rest)
Initial momentum = (m1 * v1) + (m2 * v2)
= (55 kg * 8.0 m/s) + (7 kg * 0 m/s)
= 440 kg * m/s
Total mass of person and skateboard = m1 + m2
= 55 kg + 7 kg
= 62 kg
Final velocity = (initial momentum) / (total mass of person and skateboard)
= 440 kg * m/s / 62 kg
= 7.1 m/s (rounded to one decimal place)
Therefore, the final velocity of the person and the skateboard after the interaction, assuming negligible friction, will be approximately 7.1 m/s.