A 59.5 kg person, running horizontally with a velocity of +3.90 m/s, jumps onto a 12.2 kg sled that is initially at rest.

(a) Ignoring the effects of friction during the collision, find the velocity of the sled and person as they move away.
(b) The sled and person coast 30.0 m on level snow before coming to rest. What is the coefficient of kinetic friction between the sled and the snow?

(1) use the momentum conservation law.

(2) Set the initial kinetic energy of the sled and passenger, (1/2) M V^2, equal to the work done against friction, which is
(mu,k)*M*g*X
M is the total mass
X = 30 m
9 = 9.8 m/s^2
V = initial velocity (from part a)
Solve for "mu,k", the coefficient of kinetic friction

Thanks!

To find the velocity of the sled and person as they move away, we can apply the principle of conservation of momentum. According to this principle, the total momentum before the collision is equal to the total momentum after the collision.

Let's denote the initial velocity of the person as v1 and the final velocity of the sled and person together as v2. We can set up the equation using the law of conservation of momentum:

(m1 * v1) + (m2 * 0) = (m1 * v1'') + (m2 * v2)

Where:
m1 = mass of the person (59.5 kg)
m2 = mass of the sled (12.2 kg)
v1 = initial velocity of the person (+3.90 m/s)
v1'' = final velocity of the person after the collision
v2 = final velocity of the sled and person together

Since the sled is initially at rest, its initial velocity is 0. Rearranging the equation, we can solve for v2:

(59.5 kg * 3.90 m/s) = (59.5 kg * v1'') + (12.2 kg * v2)

(232.05 kg*m/s) = (59.5 kg * v1'') + (12.2 kg * v2)

To find v2, we need to find v1''.

Since the person jumps onto the sled, we can assume an elastic collision. In an elastic collision, both momentum and kinetic energy are conserved. Therefore, we can set up an equation using conservation of kinetic energy.

Initially, the person has kinetic energy equal to (1/2) * m1 * v1^2, and after the collision, the person has kinetic energy equal to (1/2) * m1 * v1''^2.

Using this equation, we can solve for v1'':

(1/2) * (59.5 kg) * (3.90 m/s)^2 = (1/2) * (59.5 kg) * (v1'')^2

1139.215 J = 29.2025 kg * (v1'')^2

Dividing by 29.2025 kg and taking the square root, we find:

v1'' = sqrt(1139.215 J / 29.2025 kg)
= 7.264 m/s

Now that we have v1'', we can substitute it back into the equation for conservation of momentum to solve for v2:

(232.05 kg*m/s) = (59.5 kg * 7.264 m/s) + (12.2 kg * v2)

232.05 kg*m/s = 437.668 kg*m/s + (12.2 kg * v2)

Simplifying, we have:

12.2 kg * v2 = 232.05 kg*m/s - 437.668 kg*m/s

12.2 kg * v2 = -205.618 kg*m/s

Dividing by 12.2 kg, we find:

v2 = -16.86 m/s

Therefore, the velocity of the sled and person as they move away is approximately -16.86 m/s.

Moving on to part (b), we need to find the coefficient of kinetic friction between the sled and the snow. We can do this by setting the initial kinetic energy of the sled and passenger equal to the work done against friction during the 30.0 m of coasting on level snow.

The initial kinetic energy of the sled and passenger is given by:

Initial Kinetic Energy = (1/2) * (m1 + m2) * v2^2

Using the value of v2 obtained previously (-16.86 m/s) and the masses given, we can calculate the initial kinetic energy.

Initial Kinetic Energy = (1/2) * (59.5 kg + 12.2 kg) * (-16.86 m/s)^2
= 10273.8 J

The work done against friction is given by:

Work Done Against Friction = (μ_k) * (m1 + m2) * g * X

where:
μ_k = coefficient of kinetic friction (what we need to find)
m1 = mass of the person (59.5 kg)
m2 = mass of the sled (12.2 kg)
g = acceleration due to gravity (9.8 m/s^2)
X = distance traveled on level snow (30.0 m)

Setting the initial kinetic energy equal to the work done against friction, we have:

(1/2) * (59.5 kg + 12.2 kg) * (-16.86 m/s)^2 = (μ_k) * (59.5 kg + 12.2 kg) * (9.8 m/s^2) * 30.0 m

Simplifying, we find:

10273.8 J = (μ_k) * (71.7 kg) * (9.8 m/s^2) * 30.0 m

Dividing both sides by 2.08 * 71.7 * 9.8 * 30, we can solve for μ_k:

μ_k = 10273.8 J / (2.08 * 71.7 * 9.8 * 30)