Question

a 1kg box is placed in front of a spring that has a spring force constant of 1000 N/m. The spring is compressed 50 cm and the box is placed in front of the spring. The spring is then released and the 1kg box slides along the horizontal, frintionless surface until it collides and sticks to a 2 kg box which is initially at rest. what is the speed of the boxes after the collision?

Answer 1

speed of 1 kg block after it leaves the spring

0.5mv^2 = 0.5kx^2

0.5*1*v^2 = 0.5*1000*0.5^2

v = 15.81 m/s

let the speed of 1 kg box after collision be v1 and 2 kg be v2

by the conservation of momentum we have

m1u1 + m2u2 = m1v1 + m2v2

1*15.81 + 0 = 1v1 + 2v2

v1 + 2v2 = 15.81 ---------(1)

velocity of approach = velocity of recess

v2 - v1 = u1 - u2

v2 - v1 = 15.81 - 0

v2 - v1 = 15.81 -------(2)

adding both equations we have

3v2 = 31.62

v2 = 10.54 m/s

v1 = 10.54-15.81 = -5.27 m/s

so the 1 kg block moves backward with 5.27 m/s

and the 2 kg block moves in front with 10.54 m/s