Chapter 09, Problem 051 In Figure (1), a 3.50 g bullet is fired horizontally at

Question

Chapter 09, Problem 051 In Figure (1), a 3.50 g bullet is fired horizontally at two blocks at rest on a frictionless table. The bullet passes through block 1 (mass 1.36 kg) and embeds itself in block 2 (mass 1.95 kg). The blocks end up with speeds vi 0.510 m/s and v2 1.35 m/s (see Figure (2). Neglecting the material removed from block 1 by the bullet, find the speed of the bullet as it (a) leaves and (b) enters block 1. Frictionless Units (a) Number (b) Number Click if you would like to Show Work for this question: Units Open Show Work

Explanation / Answer

Let
uB the initial velocity of the bullet (unknown).
u1 the initial velocity of block 1 =0 m/s
u2 the initial velocity of block 2 = 0 m/s
v1 the final velocity of block 1 = 0.51 m/s
v2 the final velocity of block 2 =1.35 m/s
vB1 the velocity of bullet after exiting block 1 (unknown).
mB the mass of the bullet (3.5g=0.0035kg).
m1 the mass of block 1 = 1.36 kg
m2 the mass of block 2 = 1.95 kg

Total Initial Momentum i.e momentum before the bullet hits block 1
pi=mB×uB
Total Final Momentum after the bullet has struck block 2
pf=m1×v1+(m2+mB)×v2

From the conservation of momentum pi=pf.
mB×uB=m1×v1+(m2+mB)×v2

Rearrange this for the initial velocity of the bullet:
uB=m1×v1+(m2+mB)×v2 /mB
=1.36×0.51+(1.95+0.0035)×1.35 /.0035=951.7 m/s
uB=951.7m/s

Thus the velocity of bullet as it enters the block is uB=951.7m/s

Next write an expression for the total momentum after the bullet has left block 1 but before it enters block 2:
p1=mB×vB1+m1×v1

The principle of conservation of momentum requires that pi=p1 also.
mB×uB=mB×vB1+m1×v1

Rearrange for the velocity of the bullet after it leaves block 1 (vB1):
vB1=mB×uBm1×v /mB
=0.0035×951.71.36×0.510/.0035=753.5 m/s
vB1=753.5 m/s

The velocity of the bullet as it leaves block 1 is 753.5m.s1.

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