A 5 kg ball traveling to the right at 5 m/s collides with a 4 kg ball at rest. After the collision, the 4 kg ball is traveling to the right at 5.556 m/s. What is the velocity of the ball on the left after the collision?
A motor spins a disk (Idisk = 24.4 kg*m2) with a torque of 39 N*m, and a 10 N frictional force acts on the disk at a radius of 2 m. Attached to the disk at a radius of 1.1 m is a 8 kg mass. What is the magnitude of the angular acceleration in rad/s2 of the system?
here,
1)
mass of ball 1 ,m1 = 5 kg
mass of ball 2 , m2 = 4 kg
initial speed of ball 1 , u1 = 5 m/s
final speed of ball 2 , v2 = 5.556 m/s
let the velocity of the ball 1 be v1
using conservation of momentum
m1 * u1 = m1 * v1 + m2 * v2
5 * 5 = 4 * 5.556 + 5 * v1
solving for v1
v1 = 0.552 m/s
the velocity of the ball 1 after the collison is 0.552 m/s to the right
2)
the moment of inertia , I = 24.4 kg.m^2
torque , T = 39 N.m
frictional force , ff = 10 N at r1 = 2 m
mass , m = 8 kg is r2 = 1.1 m
the magnitude of angular acceleration , alpha = ( T - ff * r1) /(I + m * r2^2)
alpha = ( 39 - 10 * 2)/(24.4 + 8 * 1.1^2)
alpha = 0.56 rad/s^2
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