A rock with a mass of 0.2 kg is thrown onto a platform with mass 3kg at an angle

Question

A rock with a mass of 0.2 kg is thrown onto a platform with mass 3kg at an angle of 20degrees below the horizontal and a speed of 10m/s. The platform is initially at rest on a frictionless surface. Immediately after the collision however, the platform and the rock slide across another horizontal surface with friction and comes to a stop after 4seconds.

Note: the platform gradually covers the portion of the floor that offers friction, and therefore the normal force responsible for the friction increases!!!

1.) Calculate the magnitude of the friction force(assuming that rock gets stuck immediately to the VERY FRONT of the platform when it hits it)

2.) Calculate normal force by the front end of the platform on the rock, assuming rock's collision with the platform lasted 0.2 seconds.

I know how to do it with newton's second law, can somebody show me how to do it with impulse momentum approach????

Explanation / Answer

let

m1 = 0.2 kg, u1 = 10*cos(20) = 9.34 m/s

m2 = 3 kg, u2 = 0

m1*u1 = (m1+m2)*V

v = m1*u1/(m1+m2)

= 0.2*9.34/(0.2+3)

= 0.5873 m/s

final velosity is zero.

we know, impulse = chnage in momentum

F*t = (m1+m2)*v

F = (m1+m2)*v/t

= (0.2+3)*0.5873/4

= 0.47 N

2) Fy*t = m1*10*sin(20)

Fy = 0.2*10*sin(20)/0.2

= 3.42 N

N = Fy + m1*g

= 3.42 + 0.2*9.8

= 5.38 N

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