Question

Physics: Work and Energy/Momentum

A 6.00 Kg mass initially at rest slides down a ramp which makes an angle of $37 ^{\circ}$ with horizontal at a height of 5.0 m above the bottom of the slide and hits a 10.00 Kg mass which is at rest, after collision, the 6.00 Kg mass bounces back with a velocity of 0.60 m/s and 10.0 kg mass slides forward on level. If the coefficient of kinetic friction for the incline is 0.30 and level is 0.20 calculate: (a) The stopping distance for 10.00kg mass.(b) How long did it take to stop?

Answer 1

a)
let m1 = 6 kg

m2 = 10 kg

acceleration of block on incline, a = g*sin(37) - mue_k*g*cos(37)

= 9.8*sin(37) - 0.3*9.8*cos(37)

= 3.55 m/s^2

length of inline, d = h/sin(37) = 5/sin(37) = 8.31 m

speed of the m1 at the bottom, u1 = sqrt(2*a*d)

= sqrt(2*3.55*8.31)

= 7.68 m/s

ater the collsion, v1 = -0.6 m/s

let V is the speed of the 10 kg block after the collsion,

Apply conservation of momentum

m1*u1 = m1*v1 + m2*V

m2*V = m1*u1 - m1*v1

V = (m1*u1 - m1*v1)/m2

= (6*7.68 - 6*3.55)/10

= 2.48 m/s

acceleration of second block, a = -g*mue_k

= -9.8*0.2

= -1.96 m/s^2

distance trvelled before stopping, d = (vf^2 - vi^2)/(2*a)

= (0^2 - 2.48^2)/(2*(-1.96))

= 1.57 m <<<<<<<<<<<<<<-------------------Answer

2) timtaken, t = (vf - vi)/a

= (0 - 2.48)/(-1.96)

= 1.27 s <<<<<<<<<<<<<<-------------------Answer