Question

A 58.4-kg person, running horizontally with a velocity of +3.95 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. (Indicate the direction with the sign of your answer.)
m/s

(b) The sled and person coast 31.5 m on level snow before coming to rest. What is the coefficient of kinetic friction between the sled and the snow?

Answer 1

here,

the mass of person , m1 = 58.4 kg

the mass of sled , m2 = 12.2 kg

the initial speed of person , u1 = 3.95 m/s

a)

let the velocity of sled and person as they move together be v

using conservation of momentum

m1 * u = (m1 + m2) * v

58.4 * 3.95 = (58.4 + 12.2) * v

solving for v

v = 3.27 m/s

the velocity of sled and person as they move together is 3.27 m/s

b)

the distance traveled before stopping , s = 31.5 m

the magnitude of acceleration , a = v^2 /(2s)

a = 3.27^2 /(2*31.5) m/s^2 = 0.169 m/s^2

the coefficient of kinetic friction , uk = a /g

uk = 0.169 /9.81 = 1.73 * 10^-2