Question

Question 2 (20 marks): A 1 kg ball A is traveling horizontally at 20 m/s when it strikes a 10 kg block B that is at rest. If the coefficient of restitution between A and B is e = 0.6, and the coefficient of kinetic friction between the plane and the block is uy = 0.4, determine: 1. speed of the ball and block after the collision 2. the time for the block to stop sliding 3. the distance the block slides on the plane before it stops sliding. Figure 2.

Answer 1

mo-lokg mA = 1kg UA = 20m/s UB=0 After collesion 1. momentum conserved Maua+MBUR = MAVA + more 9x20 + loxo = 1X VAT lovu 7 VAT love = 20 - lig 2. Energy conservation. 6x146203 + 1x1066032 - LXIX VP+ bxbox V You = VA + Love - (lig 20 3. Cofficient of restitution CR = u - => 0*6 = 7 VE-VA=12 - from equ (igeiii) Vo-12 + love=20 27 11 16 = 32 2 Vo = 32 mis VA = V8 - 12 => VA = 32 -12 = -100 mis Part I Answers are vele speed of Ball = VA = -10om/s speed of Block = VB = 32 mis 1

Part-2 Ng=3%m's W = 0 from newton's equation of mosion V = wt at from FBD of Block & x = Mafn 7 x = 0.4x10x10 auon t = -ma - 40 - 10x0 = 40 * a --4 m/2 (retardation) Part 3 from equation of motion 0 = 32 + (-4)xt =) 4t = 2 3) t = Sesonds distance travelled by block before block stop from Equation of motion v² = u²+ 2as (032 = ( 32 ) + 2(-4)x 8 = 8 = -lores (12)x(+8) G= 128 m here 's' is distance covered 121
Mainly concepts involved in this question are:

1. Conservation of momentum in collision

2. Conservation of Energy in Collision

3. Newton's equations of motion