Question

If a crate of mass 5 kg is shot across a flat, frictionless surface with initial speed of 10 m/s towards a 0.25 m long region with a kinetic coefficient of friction equaling 0.10 , after passing through this region, it collides elastically with a second crate. The second crate has a mass of 0.5 kg. The second crate slides up a frictionless ramp 30 degrees off the horizontal. How high off the ground does the second crate travel before stopping?

Answer 1

here,

the mass of crate , m1 = 5 kg

initial velocity of 1 , u1 = 10 m/s

s = 0.25 m

coefficient of kinetic friction , uk = 0.1

the acceleration of friction , a = - uk * g

a = - 0.1 * 9.81 m/s^2 = - 0.981 m/s^2

let the velocity after crossing this friction be v1

using third equation of motion

v1^2 - u1^2 = 2 * a * s

v1^2 - 10^2 = 2 * (-0.981) * 0.25

v1 = 9.9 m/s

mass of second crate , m2 = 0.5 kg

let the vloeicty after the collison be v1' and v2'

using conservation of momentum

m1 * v1 = m1 * v1' + m2 * v2'

5 * 9.9 = 5 * v1' + 0.5 * v2' ....(1)

and

using conservation of kinetic energy

0.5 * m1 * v1^2 = 0.5 * m1 * v1'^2 + 0.5 * m2 * v2'^2

5 * 9.9^2 = 5 * v1'^2 + 0.5 * v2'^2 ....(2)

from (1) and (2)

v1' = 8.1 m/s

v2' = 18 m/s

the height traveled , h2 = v2'^2 /(2g)

h2 = 18^2 /(2*9.81) m

h2 = 16.5 m