A bullet with mass 4.98 g is fired horizontally into a 2.043-kg block attached t

Question

A bullet with mass 4.98 g is fired horizontally into a 2.043-kg block attached to a horizontal spring. The spring has a constant 6.50 102 N/m and reaches a maximum compression of 6.44 cm.

a) Find the initial speed of the bullet-block system.


(b) Find the speed of the bullet.

Explanation / Answer

according to law of conservation of momentum, mu1+Mu2=(m+M)V 0.00498*u1+0=(2.043+0.00498)V---->A kinetic energy lost by bullet-block system=potential energy gained by spring 1/2(m+M)V^2=1/2Kx^2 V^2=6.5*10^2*(6.44*10^-2)^2/(2.043+0.00498) =2.695784/2.04798 =1.316 V=1.147m/s a)initial speed of bullet-block system=1.147m/s b)speed of bullet,u1=(2.043+0.00498)*1.147/0.00498 =471.69m/s

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