You shoot a 0.0050-kg bullet into a 2.0-kg wooden block at rest on a horizontal

Question

You shoot a 0.0050-kg bullet into a 2.0-kg wooden block at rest on a horizontal surface (Figure 1) . After hitting dead center on a hard knot that runs through the block horizontally, the bullet pushes out the knot. It takes the bullet 1.0 ms to travel through the block, and as it does so, it experiences an x component of acceleration of -4.9 × 105 m/s2. After the bullet pushes the knot out, the knot and bullet together have an x component of velocity of +10 m/s. The knot carries 10% of the original inertia of the block. A- What is the initial velocity of the bullet? Express your answer with the appropriate units. B- Using conservation of momentum, compute the final velocity of the block after the collision. Express your answer with the appropriate units. C- Calculate the initial kinetic energy of the block-knot-bullet system. Express your answer with the appropriate units. D- alculate the final kinetic energy of the block-knot-bullet system. Express your answer with the appropriate units.

Explanation / Answer

Given that

The acceleration of the bullet (a) =-4.9 × 105 m/s2

The time (t) =1.0 ms =1*10-3s

A)

The knot and bullet together have an x component of velocity of (v) =+10 m/s.

Now using the equation v =u+at

then the initial velocity of the bullet is (u) =v-at =10+(4.9 × 105 m/s2)(10-3s)=500m/s

B)

Now from the conservation of momentum

m1u1+m2u2 =(0.10*2+0.0050)*10+0.90*2*v

0.0050*500+0 =(0.10*2+0.0050)*10+0.90*2*v

then 1.8v =2.5-2.05=0.45====>v=0.25m/s

c)

The initial kinetic energy of thenlock-knot-bullet suystem is KEinitial =(1/2)m1u12 =0.5*0.0050*(500)2 =625J

D)

The final kinetic energy of thenlock-knot-bullet suystem is KEinitial =(1/2)[(0.0050+0.1*2)(10)2 +(0.9*2)*(0.25m/s)2] =[20.5+0.1125]/2=10.306J

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