Question

Block B of mass 10.0 kg is placed in contact with an unstretched spring on a horizontal, frictionless surface. The other end of the spring is attached to a fixed support. Block A with a mass of 4.00 kg is moving with a speed of 20.0 m/s when it collides with and sticks to B. (a) What is the speed of the combined blocks after the collision? The blocks compress the spring 2.60 m before coming to rest momentarily. (b) What is the force constant of the spring? (c) How long after the collision does the spring reach this maximum compression first?

Answer 1

a)

apply linear momentum conservation

Pai +Pbi = Pabf

4*20 + 0 = ( 4+10)*vf

vf =80/14 = 5.7142 m/s

so speed of combined blocks =5.71428571 m/s

b)

assuming there is no friction force on the ground

kinetic energy of two block = goes into stored energy in spring

1/2*(10+4)*5.714^2 = 1/2*k*2.6^2

k =  (10+4)*5.714^2 /2.6^2 =

k≈67.6179 N/m

c)

just before collision spring was in its natural lenght so it was equilibrium position so from equilibrium to maximum compression

time taken = t = T/4  

m = 10+4 =14 kg

T = 2pi *sqrt( m/k) = 2*pi*sqrt( 14 / 67.6179 ) = 2.8589927

so t= 2.8589927/4 = 0.714748175 sec answer

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