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maks197457 [2]
3 years ago
10

A 4.10 g bullet moving at 837 m/s strikes a 820 g wooden block at rest on a frictionless surface. The bullet emerges, traveling

in the same direction with its speed reduced to 467 m/s. (a) What is the resulting speed of the block? (b) What is the speed of the bullet-block center of mass?
Physics
1 answer:
atroni [7]3 years ago
6 0

Answer:

(a) 1.85 m/s

(b) 4.1 m/s

Explanation:

Data

  • bullet mass, Mb = 4.10 g
  • initial bullet velocity, Vbi = 837 m/s
  • wooden block mass, Mw = 820 g
  • initial wooden block  velocity, Vwi = 0 m/s
  • final bullet velocity, Vbf = 467 m/s

(a) From the conservation of momentum:

Mb*Vbi + Mw*Vwi = Mb*Vbf + Mw*Vwf

Mb*(Vbi - Vbf)/Mw = Vwf

4.1*(837 - 467)/820 = Vwf

Vwf = 1.85 m/s

(b) The speed of the center of mass speed is calculated as follows:

V = Mb/(Mb + Mw) * Vbi

V = 4.1/(4.1 + 820) * 837

V = 4.1 m/s

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a bullet moving with a velocity of 100m/s pierce a block of wood and moves out with a velocityof 10 m/s.if the thickness of the
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The emerging velocity of the bullet is <u>71 m/s.</u>

The bullet of mass <em>m</em> moving with a velocity <em>u</em>  has kinetic energy. When it pierces the block of wood, the block exerts a force of friction on the bullet. As the bullet passes through the block, work is done against the resistive forces exerted on the bullet by the block. This results in the reduction of the bullet's kinetic energy. The bullet has a speed <em>v</em> when it emerges from the block.

If the block exerts a resistive force <em>F</em> on the bullet and the thickness of the block is <em>x</em> then, the work done by the resistive force is given by,

W=Fx

This is equal to the change in the bullet's kinetic energy.

W=Fx=\frac{1}{2} m(u^2-v^2)......(1)

If the thickness of the block is reduced by one-half, the bullet emerges out with a velocity v<em>₁.</em>

Assuming the same resistive forces to act on the bullet,

F(\frac{x}{2} )=\frac{1}{2} m(u^2-v_1^2)......(2)

Divide equation (2) by equation (1) and simplify for v<em>₁.</em>

\frac{\frac{Fx}{2} }{Fx} =\frac{(u^2-v_1^2)}{(u^2-v^2)} \\\frac{100^2-v_1^2}{100^2-10^2} =\frac{1}{2} \\v_1^2=5050\\v_1=71.06 m/s

Thus the speed of the bullet is 71 m/s


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