Question

1) A turntable is found to go from an angular speed of 24rad/s to 21 rad/s in a time of 10s, if it starts at 23rad/s how long until it comes to rest? 2) The ball is dropped from a height of 1.3 m and has a mass of 2kg. what is the change in momentum of the ball on impact with the ground in kg m/s? 3) Two blocks of mass m and speed v moving in the same direction collide with a third block of mass m at rest and they stick together. If the surface is frictionless, what is the final speed of the three blocks?

Answer 1

Ans

   Given
1)
turn table angular speed w1 = 24 rad/s, w2 = 21 rad/s, in time t = 10 s

the angular acceleration is W2 = W1+alpha *t ==> alpha = w2-w1/t = 21-24/10 =-3/10 = - 0.3 rad/s2

now the time taken for table , come to rest from 23 rad/s is t = 0-23/(-0.3)= 76.67 s

2)
mass of ball m = 2 kg, height from where it dropped is h = 1.3 m

coservation of enegy mgh = 0.5 mv^2 ==> v= sqrt(2gh)

before impact v = -sqrt(2gh)

after impact + sqrt(2gh)

AND the momentum of the ball upon striking the ground is m*sqrt(2gh) = 2*sqrt(2*9.8*1.3) kg m/s = 10.096 kg m/s

so chang in momentum is dP = ((sqrt(2gh) -(-sqrt(2gh))) = 2*sqrt(2gh)= 2*10.096 = 20.192 kgm/s = 20.2 kg m/s

3) by conservation of momentum

m*u+m*U = (3m)v
v = 2m*u/3m
V = 2/3*u
so the final speed of the three block is 2/3 rd of initial speed