Your friend braked the motorcycle and brought it to a complete standstill within 10 seconds. Can you determine the rate of change of momentum for this? What is the amount of force applied to the motorcycle?

 To determine the rate of change of momentum and the amount of force applied to the motorcycle, we can use the concept of impulse and Newton's second law of motion.

Rate of Change of Momentum

The initial momentum (${𝑝}_{\text{initial}}$) of the motorcycle and rider is given by:

${𝑝}_{\text{initial}}=6000\text{kg}\cdot \text{m/s}$

The final momentum (${𝑝}_{\text{final}}$) when the motorcycle comes to a complete standstill is:

${𝑝}_{\text{final}}=0\text{kg}\cdot \text{m/s}$

The change in momentum ($\mathrm{\Delta }𝑝$) is:

$\mathrm{\Delta }𝑝={𝑝}_{\text{final}}-{𝑝}_{\text{initial}}$ $\mathrm{\Delta }𝑝=0\text{kg}\cdot \text{m/s}-6000\text{kg}\cdot \text{m/s}$ $\mathrm{\Delta }𝑝=-6000\text{kg}\cdot \text{m/s}$

The rate of change of momentum is the change in momentum divided by the time interval over which the change occurs. Given that the time interval ($\mathrm{\Delta }𝑡$) is 10 seconds:

$\text{Rate of change of momentum}=\frac{\mathrm{\Delta }𝑝}{\mathrm{\Delta }𝑡}$ $\text{Rate of change of momentum}=\frac{-6000\text{kg}\cdot \text{m/s}}{10\text{s}}$ $\text{Rate of change of momentum}=-600\text{kg}\cdot {\text{m/s}}^2$

The negative sign indicates that the momentum is decreasing.

Amount of Force Applied

According to Newton's second law, the force ($𝐹$) applied is equal to the rate of change of momentum:

$𝐹=\text{Rate of change of momentum}$ $𝐹=-600\text{N}$

The magnitude of the force applied to bring the motorcycle to a standstill is 600 N. The negative sign indicates that the force is applied in the direction opposite to the motion (i.e., it is a braking force).

So, the rate of change of momentum is $-600\text{kg}\cdot {\text{m/s}}^2$, and the force applied is 600 N in the direction opposite to the motorcycle's motion.