Question

a block of mass m1=3kg, starting from rest at a height h, slides down a frictionless ramp along a frictionless surface before it collides and sticks to a second block m2= 4kg. the second block is attatched to a spring with a spring constant of 252N/m which is fixed to a rigid wall at its other end. after the collision the blocks compress the spring 0.30 meters before they momentarily come to rest. from what height h did the first block begin to slide down the ramo

PLEASE HELP DUE IN AN HOUR

Answer 1

here,

the mass 1 , m1 = 3 kg

m2 = 4 kg

the spring constant , K = 252 N/m

the compression in the spring , x = 0.3 m

let the velocity of combined mass be v

using conservation of energy

0.5 * ( m1 + m2) * v^2 = 0.5 * K * x^2

0.5 * (3 + 4) * v^2 = 0.5 * 252 * 0.3^2

solving for v

v = 1.8 m/s

let the initial velocity of m1 before the collison be u

using conservation of momentum

m1 * u = (m1 + m2) * v

3 * u = (3 + 4) * 1.8

u = 4.2 m/s

the initial height of mass 1 , h = u^2 /(2g)

h = 4.2^2 /(2*9.81) m

h = 0.9 m