89. A 5.00-g bullet mov-AMT ing withM speed of v; = 400 m/sis fired into and passesVian initial1.00-kgblock as shown in Fig-ure P9.89. The block,initially at rest on africtionless, horizontalsurface, is connectedto a spring with forceN/m.throughaFigure P9.89900constantThe block moves d = 5.00 cm to the right after impactbefore being brought to rest by the spring. Find (a) thespeed at which the bullet emerges from the block and(b) the amount of initial kinetic energy of the bulletthat is converted into internal energy in the bullet-block system during the collision.

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Asked Dec 5, 2019
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89. A 5.00-g bullet mov-
AMT ing with
M speed of v; = 400 m/s
is fired into and passes
Vi
an initial
1.00-kg
block as shown in Fig-
ure P9.89. The block,
initially at rest on a
frictionless, horizontal
surface, is connected
to a spring with force
N/m.
through
a
Figure P9.89
900
constant
The block moves d = 5.00 cm to the right after impact
before being brought to rest by the spring. Find (a) the
speed at which the bullet emerges from the block and
(b) the amount of initial kinetic energy of the bullet
that is converted into internal energy in the bullet-
block system during the collision.
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89. A 5.00-g bullet mov- AMT ing with M speed of v; = 400 m/s is fired into and passes Vi an initial 1.00-kg block as shown in Fig- ure P9.89. The block, initially at rest on a frictionless, horizontal surface, is connected to a spring with force N/m. through a Figure P9.89 900 constant The block moves d = 5.00 cm to the right after impact before being brought to rest by the spring. Find (a) the speed at which the bullet emerges from the block and (b) the amount of initial kinetic energy of the bullet that is converted into internal energy in the bullet- block system during the collision.

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Expert Answer

Step 1

Consider the bullet’s mass be m, the block’s mass be M, the initial speed of the block be vbi, and the initial speed of the block be vbf.

 

The given values are,

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1 kg m = 5.00 g = 5.00 g 1000 g = 0.005 kg M =1.00 kg v, = 400 m/s k : = 900 N/m = 0.05 m d = 5.00 cm = 5.00 cm 100 cm

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Step 2

(a)

 

After the bullet enters the block, the kinetic energy of the block will conserve into the spring.

 

Apply energy conservation on spring block system after the bullet enters the block, and solve for the final speed of the block.

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kd Mv kd Vof Vif = (900 N/m)(0.05 m)´ 1.00 kg Jjasosmy 1 kg m/s? (0.05 m)¯ (900 N/m' 1N 1.00 kg = 1.50 m/s

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Step 3

Apply momentum conservation on the bullet-block system, and solve f...

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mv, + Mv = mv; + Mvg (0.005 kg)(400 m/s) + M (0) = (0.005 kg)v, +(1.00 kg)(1.50 m/s) = mv, + Mv, bi (2.00 kg - m/s)- (1.50 kg - m/s) = (0.005 kg)v; %3D = 100 m/s

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