6. How fast is a speeding bullet? The figure below shows a simple form of a ball

Question

6. How fast is a speeding bullet? The figure below shows a simple form of a ballistic pendulum, which is a system for measuring the speed of a bullet. The bullet, with mass ms. is fired into a block of wood with mass m, suspended like a pendulum. a. The bullet makes a completely inelastic collision with the black, becoming embedded in it. After the impact, the black swings up to a maximum height h. What is the initial speed v of the bullet? What becomes of its initial kinetic energy? Give the initial speed in terms of m, m, h, and any other physical constants necessary b. f m 5.00 g. m- 2.00 kg, h- 3.00 cm, what is the initial speed vof the bullet? c. Using the values given in (b), what is the total kinetic energy of the system just before and just after impact? Explain why your answer makes sense. RHFORECXLISON mo ly TOP OF SWING AFTER COLLISION

Explanation / Answer

a)From conservation of energy:
KE = PE

1/2 (mB + mW) vf^2 = (mB + mW) g h

vf = sqrt ( 2 g h)

From conservation of momentum

Pf = Pi

(mB + mW)vf = mB vi

vi = (mB + mW)vf / m1

putting vf = sqrt (2 g h), we get

vi = (mB + mW)sqrt ( 2 g h)/m1

Hence, vi = (mB + mW)sqrt ( 2 g h)/m1

b)vi = (mB + mW)sqrt ( 2 g h)/m1

vi = (0.005 + 2) sqrt (2 x 9.81 x 0.03)/0.005 = 307.67 m/s

Hence, vi = 307.67 m/s

c)KEi = 1/2 mB vi^2

KEi = 0.5 x 0.005 x 307.55^2 = 236.65 J

vf = sqrt (2 g h) = sqrt (2 x 9.81 x 0.03) = 0.77 m/s

KEf = 0.5 x (0.005 + 2) x 0.77^2 = 0.59 J

Hence, Kei = 236.65 J and KEf = 0.59 J

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