please help me these questions. A ball of mass mB = 1/9 [kg] is fired directly a

Question

please help me these questions.

A ball of mass mB = 1/9 [kg] is fired directly and centrally at a target of mass mr = 1 [kg] which is resting on a flat horizontal floor. When the ball strikes the target, the ball has a purely horizontal speed vBi = 320 [m/s]. The coefficient of restitution between the ball and the target is e = 1/4 and the coefficient of friction between the target and the floor is mu = 2/1. Taking g = 10 [m/s2], find the velocity of the ball immediately after the collision, the velocity of the target immediately after the collision, and the time it takes the target to come to rest again. Do all 3 problems. Closed book, closed notes, no calculators. Read the instructions carefully. Write your name on every page. Feel free to use the back side of each page, but keep your work neat and readable. Work must be shown for full credit. A pendulum consisting of a massless string of length R and point mass m is released from rest at theta 0 = pi / 6. Taking g = 10 [m/s2], find the velocity of the pendulum bob when it first arrives at a position directly under the pivot point O by both of the following methods: traditional force analysis, and using the Work/Energy Theorem.

Explanation / Answer

2)e=vb-va/(ua-ub)

=1/4=(vT-vb)/(320-0) as ub=320 and uT=0

=vT-vb=80

Also applying conservation of momentum

mb*ub+mT*0=mb*vb+mT*vT

1/9*320=1/9*vb+1*vT

vb+9vT=320

vT=40m/s and vb=-40m/s

a)velocity of ball imidiately after the collision vb=-40m/s

b)velocity of target imiduately after the collision vT=40 m/s

c)applying conservation of energy i.e. Work done by frictional force=Change in its kinetic energy

1*10*2/7*x=1/2*1*40^2

x=280 m

now v^2=u^2+2*a*x

a=-(40)^2/(2*280)

=-2.85 m/s^2

also v=u+at

t=14 s ANS

Second question Answer

1) Net centripital force is provided by the net external force

at bottom we have N-mg=mv^2/R

v=sqrt(R*(N-mg)/m)

b)By conservation of energy

change in potential energy=change in kinetic energy

m*g*R*(1-cospi/3)=1/2*m*v^2

v=sqrt(2*m*g*R*(1-cospi/3)/m) ans

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