The speed of a billet can bee assured by firing a gun at a block of soft wood th

Question

The speed of a billet can bee assured by firing a gun at a block of soft wood that allows the bullet to embed itself inside and analyzing the subsequent motion. A light rod of length l is 50.0 cm if attached at one end to a frictionless hinge and to a wooden block of mass M is 100.0 g at the other end. The rod can swing freely in a vertical plane. A bullet with mass m is 30.0 g is fired at the block embedding itself in it. The rod swing up from the initial vertical position to an angle theta is 20.0 deg. Find the velocity of the bullet as it exits theuzzle of the gun? The speed of a billet can bee assured by firing a gun at a block of soft wood that allows the bullet to embed itself inside and analyzing the subsequent motion. A light rod of length l is 50.0 cm if attached at one end to a frictionless hinge and to a wooden block of mass M is 100.0 g at the other end. The rod can swing freely in a vertical plane. A bullet with mass m is 30.0 g is fired at the block embedding itself in it. The rod swing up from the initial vertical position to an angle theta is 20.0 deg. Find the velocity of the bullet as it exits theuzzle of the gun?

Explanation / Answer

let,

length of the rod l=50cm

mass of the block M=100g

mass of the bullet m=30g

theta=20 degrees,

and

initial speed of the bullet is u1=u

after collision,

speed of the bullet and block is v

by using law of conservation momentum,

m*u1+M*u2=(m+M)*v

m*u+0=(m+M)*v

u=((m+M)/m)*v ----(1)

by using law of conservatio energy,

K.E=P.E

1/2*(m+M)*v^2=(m+M)*g*h

1/2*v^2=g*l(1-cos(thet))

===>

V^2=2*g*l(1-cos(thet))

V^2=2*9.8*0.5(1-cos(20))

===> V=0.77 m/sec

after collision the speed of the bullet and block is v=0.77 m/sec

from equaion (1),

u=((m+M)/m)*v

u=((30+100)/30)*0.77

u=3.33 m/sec

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