Example 9-12 depicts the falowing scenaria. In a ballstic pendulum, an chject of

Question

Example 9-12 depicts the falowing scenaria. In a ballstic pendulum, an chject of mass is fired with an intial speed , a intlai speed w, t the bob of a penduum The bob haa amass M a and is suspended by a rod of negigbie mass Ater the collision, the object and the bob sick together and swirg through an arc, eventualy gaining a he ght aining a Now we l consider same slightly dierent relaled senarios to Example 9-12 Part A Suppose a bulel of ma8s -844 gi fired inlo a balielie pendulam whose bob has a mass ef M-0675 kg thebob rises ko a heighl of 0.118 m, whal was the inifial speed of the bulet m/s My Answers Give Up Part B what was the spoed of the buleI.bob combination immedidely dn the colison took plata? Express your answer using three significant igures m/s My Answrs Give up Part C Reta back to example 3 12 bul t ith a mass of m 8.10 and an ial s ad of t 320 m si fied into a balistic pand lun wh t masa must the bat have iftha2 buk-bob ca tiation is to rise to a ma imum height on 125 m an the e ion? Express your answer using two signiticant tgures 4 kg

Explanation / Answer

from momentum conservation principle

momentum before collision = momentum after collision

m*vi = (M+m)*vf

vf = m*vi/(m+M)

after collision the combined mass (m + M) moves to a height

KE of combined after collision = potential energy of combined mass at height h

(1/2)*(m+M)*vf^2 = (m+M)*g*h

vf^2 = g*h

vf =sqrt(2gh)

m*vi/(M+m) = sqrt(2gh)

vi = (M+m)/m*sqrt(2gh)

vi = (1 + M/m) *sqrt(2gh)

part (A)

vi = (1 + 0.675/0.00644)*sqrt(2*9.8*0.118)

vi = 161 m/s <<<<-----ANSWER

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part B

vf = m*vi/(m+M)

vf = sqrt(2gh) = sqrt(2*9.8*0.118) = 1.52 m/s <<<-----ANSWER

====================

part (c)

vi = (1 + M/m)*sqrt(2g*h)

320 = (1 + M/0.0081)*sqrt(2*9.8*0.125)

M = 1.65 kg <<<<<-----------ANSWER

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