Block A, of mass 180.0g, is initially at the end of a compressed spring. The spr

Question

Block A, of mass 180.0g, is initially at the end of a compressed spring. The spring of force constant 240.0N/m is compressed 15.0cm from its equilibrium position. The block is released, and it moves ona surface with coefficient of friction 0.360. When block A moves 10.0cm from the spring's equilibrium position, it collides elastically with block B, of mass 125g. The coefficient of friction between the surface and block B is also 0.360.

(a) What is the speed of block A just before it hits block B?

(b) What is the speed of block B due to the collision with block A?

(c) How far does block B go before it stops?

Explanation / Answer

a)Energy conservation upto just before colllision gives

0.5kx^2=0.5mv^2 +0.36mg(x+0.1){included work done by friction for x+0.1=0.25 m}

So

0.5*240*0.15^2=0.5*0.18*v^2+0.36*0.18*9.8*0.25

So

v=5.477m/s

b)0.18*5.477=0.125v'+0.18v'' (momentum conservation)

and

v'-v''=5.477 (as e=vel. of separation/vel. of approach=1 for elastic collision)

So

v''=0.987m/s (vel of A towards right)

v'=6.465 m/s (vel of B towards right)

a=-0.36*9.8=-3.528m/s^2

v^2-u^2=2as

So

s=(0-6.465^2)/(-2*3.528)=5.92m

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