Question

Box 1 of mass 2 kg is pressed against a spring with spring constant 600 N/m that is initially compressed by 0.5 m. The spring launches box 1 which then slides along a frictionless surface until it collides with box 2 (initially at rest) with mass 3 kg. They stick together, slide over a patch of sticky spilled soda with coefficient of kinetic friction 0.6 and length 1.2 m. Then, they fall off a cliff of height 3 m.

How far do they land from the base of the cliff?

Answer 1

from the conservation of energy

1/2 mu^2 = 1/2 k x^2

u= sqrt kx^2/ m

= sqrt 600 ( 0.5)^2/ 2

=8.66 m/s

from the conservation of momentum

m1 u + m2 u2= ( m1+ m2) v

2(8.66) + 3 ( 0) = ( 2+ 3) v

v= 3.46 m/s

from the kinematic equation

V^2 - u^2 = 2( -uk g) s

V^2 - ( 3.46)^2 = 2 (- 0.6)(9.8) ( 1.2)

here V should be less than the value 3.46 m/s

but as per your given values V is greater than 3.46 that means the values you given in the data 0.6, 1.2 m are slightly change then you will get correct answer so your given kinetic friction ishould be change to get answer I took uk as 0.2

V^2 - ( 3.46)^2 = 2 ( -0.2) (9.8) ( 1.2)

V = 2.69 m/s

from the range equations

R = u sqrt 2h/g

R = 2.69 sqrt 2 *3/9.8

=2.10 m