A block of mass 0.1 kg is attached to a spring of spring constant 18 N/m on a fr

Question

A block of mass 0.1 kg is attached to a spring of spring constant 18 N/m on a fric- tionless track. The block moves in simple har- monic motion with amplitude 0.29 m. While passing through the equilibrium point from left to right, the block is struck by a bullet, which stops inside the block. The velocity of the bullet immediately before it strikes the block is 64 m/s and the mass of the bullet is 1.8 g.

If the simple harmonic motion after the collision is described by x = B sin( t + ), what is the new amplitude B?
Answer in units of m.

Explanation / Answer

= ( k/ m ) ;

= ( 18/ 0.1 ) = 13.416

velocity of block at equilibrium position ,

v = A ;

v = 13.416 * 0.29

v= 3.890 m /s ;

conservation of momentum at equilibrium point ;

Mu1 + m u2 = (M + m )V ;

0.1 * 3.890 + 1.8e-3 * 64 = ( 0.1 + 1.8e-3 ) V ;

V = 4.953 m / s ;

conservation of mechanical energy after collision ,

1/2 ( M + m ) V^2 = 1/2 k B^2;

( M + m ) V^2 = k B^2;   

( 0.1 + 1.8e-3 ) * 4.953^2 = 18 * B^2 ;

B = 0.3725 m <---------------ans

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