Question

2.(a) Ball A is released from rest. It collides with the stationary ball B with a velocity 3.2 m/s; immediately after the collision ball A travels in the same direction with velocity 2.3 m/s.   

Ball A has mass 0.26 kg; ball B has mass 0.07 kg.

Calculate

(i) the velocity of ball B immediately after the collision..

(ii) the maximum height reached by ball B.

(b) A driver is travelling at a constant speed of 15.4 m/s in a 1800 kg car.

At this speed he then enters a large empty car park, and makes a U-turn, travelling in a complete half-circle of radius r.

The friction force between the tyres and the ground is 12.4 kN.

Calculate r.

Answer 1

ball A mass = 0.26 kg

ball B mass = 0.07 kg

i) during collision momentum is conserved

momentum before collision   = momentum after collision

lets take the velocity of ball B after collision be v

0.26(3.2) + 0.07(0) = 0.26(2.3) + 0.07(v)

0.07(v) = 0.26( 3.2-2.3)

v = 3.343 m/sec

ii) maximum height reached by the ball B after collision is = v*v/2*g

                                                                                             = 0.57 mts

b) mass of vehicle = 1800 kg

speed of vehicle is = 15.4 m/sec

frictional force is balanced by centripetal force

frictional force            =    m*v*v/r

12400                   = 1800* 15.4*15.4  /r

r          =   1800*15.4*15.4/12400

radius (r)   = 34.43 mts