Question

A ball of mass 60 g is dropped from a height of 3.4 m. It lands on the top of a ramp at height 1.8 m with a kinetic friction coefficient of .3. The ramp is tilted at an angle of 20 degrees. 

(a) What is the velocity of the ball at the top of the ramp?

(b) At the bottom of the ramp it collides with and sticks to a ball of mass 73 g. What is their velocity after the collision?

(c) The stuck together balls collide with a spring of spring constant 300 N/m. How much will they compress it?

(d) They then go back up the ramp. How high will they go?

Answer 1

here,

mass of ball 1 , m1 = 60 g = 0.06 kg

initial height , h1 = 3.4 m

h2 = 1.8 m

uk = 0.3

theta = 20 degree

a)

the velocity of the ball at the top of the ramp , u = sqrt(2 * g * ( h2 - h1))

u = sqrt(2 * 9.81 * ( 3.4 - 1.8))

u = 5.6 m/s

b)

the velocity of ball1 before the collison be v

using work energy theorm

work done by friction = change in mechanical energy

- uk * m * g * cos(theta) * (h2 /sin(theta)) = 0.5 * m * v^2 - 0.5 * m * u^2 - m * g * h1

- 0.3 * 9.81 * cos(20) * (1.8 /sin(20)) = 0.5 * v^2 - 0.5 * 5.6^2 - 9.81 * 1.8

solving for v

v = 6.13 m/s

let the speed after the collison be v'

using conservation of momentum

m1 * v = (m1 + m2) * v'

0.06 * 6.13 = ( 0.06 + 0.073) * v'

v' = 2.77 m/s

let the compression in the spring be x

using conservation of energy

0.5 *(m1 + m2) * v'^2 = 0.5 * K * x^2

(0.06 + 0.073) * 6.13^2 = 300 * x^2

solving for x

x = 0.13 m

the compression in the spring is 0.13 m

d)

the height above the ramp be h3

work done by friction = change in mechanical energy

- uk * (m1 + m2) * g * cos(theta) * (h3 /sin(theta)) = (m1 + m2) * g * h3 - 0.5 * K * x^2

- 0.3 * (0.06 + 0.073) * 9.81 * cos(20) * (h3 /sin(20)) = (0.06 + 0.073) * 9.81 * h3 - 0.5 * 300 * 0.13^2

h3 = 1.07 m

the height gained is 1.07 m