Two blocks are free to slide along the frictionless wooden track shown below. Th

Question

Two blocks are free to slide along the frictionless wooden track shown below. The block of mass m1 = 4.99 kg is released from the position shown, at height h = 5.00 m above the flat part of the track. Protruding from its front end is the north pole of a strong magnet, which repels the north pole of an identical magnet embedded in the back end of the block of mass m2 = 10.6 kg, initially at rest. The two blocks never touch. Calculate the maximum height to which m1 rises after the elastic collision.

________ m

Explanation / Answer

m1 = mass of first block = 4.99 kg

m2 = mass of second block = 10.6 kg

h = height = 5m

u1i = initial velocity of m1

u2i = iniyial velocity of m2 = 0 due to at rest

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from energy conservation during initial fall of m1

K.Ei + P.Ei = K,Ef + P.Ef

0 + m1*g*h = ( 1/2)*m1*V12 + 0

V1 = ( 2*g*h)1/2   --------------------------------1

now this is the initial velocity of m1 ,with which it collide with m2.

ie u1i = ( 2*g*h)1/2  

From elastic collision in 1 - dimension:

V1f = [ ( m1-m2)*u1i / m1+m2 ] + [ 2*m2*u2i / m1+m2]

here V1f = final velocity of m1 ,after collision, u2i = 0 , and by using eq 1

V1f = (m1-m2)*( 2*g*h)1/2 / (m1+m2)

  Let new height after collision be h'.

from energy conservation,

(1/2)*m1*V1f2 = m1*g*h'

h' = V1f2 / 2g

= ( (m1- m2) / m1+m2 )2 ( 2gh / 2g) [ put the vlue of Vif]

= ( 4.99- 10.6 / 4.99 + 10.6)2 * 5

= 0.647 m

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