Question

A motorcycle of mass 100 kilograms slowly rolls off the edge of a cliff and falls for three seconds before reaching the bottom of a gully. Its momentum upon hitting the ground is

Answer 1

Answer:

2940 kg m/s

Explanation:

First of all, we need to find the final velocity of the motorcycle when it hits the ground. The horizontal component of the velocity is zero, so we can just find the vertical component, which is given by:

$v_y = v_0+gt$

where

v0 =0 is the initial vertical velocity

g = 9.8 m/s^2 is the acceleration due to gravity

t = 3 s is the time of the fall

Substituting into the formula, we find

$v_y=0+(9.8 m/s^2)(3 s)=29.4 m/s$

And so, we can now find the momentum of the motorcycle upon hitting the ground, which is given by the product between the mass and the velocity:

$p=mv=(100 kg)(29.4 m/s)=2940 kg m/s$