A block of mass m 1 = 2.5 kg slides from height 10.0 meters on a frictionless su

Question

A block of mass m1 = 2.5 kg slides from height 10.0 meters on a frictionless surface and collides with a block of mass m2 = 4.0 kg.

Hint: Conservation of momentum: total Initial momentum of the system = total final momentum of the system Pi =Pf                   P = mv

Conservation of mechanical energy   Ei = Ef              E= PE + KE         PE= mg H      KE=1/2 mv2

V1f = ( m1 - m2 ) V1i/ ( m1 + m2 )       V2f = ( 2m1 ) V1i/ ( m1 + m2)

( m1 V1i + m2 V2i ) =( m1 + m2 )Vf        

Find the final velocities of the two blocks after collision if the Collison is elastic.

**show all the steps to the final answers.

Find the final velocities of the two blocks after collision if the Collison is inelastic.

***show all the steps to your final answers.

What is the maximum height to which m1 rises after the collision?

**show all the steps to your final answers.

Explanation / Answer

given m1 = 2.5 kg

h = 10 m

m2 = 4 kg

so speed of block 1 justbefore collision = u

from conservation of energy

m1gh = 0.5m1*u^2

u = 14 m/s

after collision

speed og blocks = v1, v2

from conservation of momentum

m1u = m1v1 + M2V2

2.5*14 = 2.5v1 + 4v2

a. for elastic collision

(u - 0)/(v2 - v1) = 1

14 = v2 - v1

35 = 2.5v1 + 4(14 + v1)

v1 = -3.228 m/s

v2 = 10.7719 m/s

b. for inelastic collision

v1 = v2

v1 = v2 = 5.384 m/s

c. for elastic collsision maximum height of m1 = H

0.5*v1^2 = g*H

H = 0.531 m

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