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 = 5.09 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 = 10.9 kg, initially at rest. The two blocks never touch. Calculate the maximum height to which m1 rises after the elastic collision.

in meters

Explanation / Answer

Given are

M1=5.09kg

H1=5m

M2=10.9kg

H2=?

Step 1—first we will use conservation of energy and then will find the velocity of first block before the collision

So

K1+Ui=Kf+Uf

0+mgh=1/2mv1^2+0

Mgh=1/2mv1^2

V=undrt(2gh)---1

Now since the collision is elastic so K.E and momentum both will be conserved

So

V1f=(m1-m2)v1i/(m1+m2)

Plugging in value of eqn 1 we get

V1f=(m1-m2)/(m1+m2)undrt(2gh)

Now we will use conservation of energy

1/2m1v1^2=m1gh’

H’=v1f^2/2g

=h((m1-m2)/(m1+m2)^2

=h(5.09-10.9/5.09+10.9))^2

=0.132h

=0.132*5

H’=0.660m--answer

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