A 64.0-kg person, running horizontally with a velocity of +3.76 m/s?

Question

A 64.0-kg person, running horizontally with a velocity of +3.76 m/s, jumps onto a 10.8-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?

Answer 1

a)

No external forces, therefore momentum is conserved....

m1=64Kg
m2=10.8Kg
v1=+3.76ms
v2=0 initial velocity of sled

let's assume that finally they both move together in the direction of the person.....

m1 X v1+ m2 X v2= (m1+m2) X Vf....

U can't apply energy conservation simply becuase the collision is inelastic(e=0) since both sled and person stick to each other, move together.

b)After calculating Vf from above equation.....

Since only friction is the only external force acting....

Therefore

Workdone by friction in stopping = Change in KE(after collision).

Work Friction= - k(m1+m2) g X D where D=30.0m, k=kinetic friction coeff, g=9.8m/s^2. Its -ve since Friction is acting opposite the direction of motion

Change in KE= Final KE- Initial KE
Final KE=0 since it has stopped.
Initial KE= KE after collision = 1/2(m1+m2) X Vf^2

equate these 2 to get k

Answer 2

I don't get it. Did someone turn on friction just after the person got in the sled? Or maybe this place has no static friction and only kenetic friction? Wait, are you suggesting that in the case of a sled, static friction can be neglected?

Answer 3

(a) Momentum mv is conserved, so m(1) v(1) = m(2) v(2);

64.0 * 3.76 / 10.8 = v(2)

(b) You do this one.