1) A block (m1) slides from rest down a frictionless slide from a height h and c

Question

1) A block (m1) slides from

rest down a frictionless slide

from a height h and collides

with a second block (m2=2m1)

initially at rest. The collision

is elastic. After the collision

area is a rough part of the

horizontal surface.

Your ultimate goal is to provide a formula for how far m2 will travel on the rough surface before

stopping.

(a) Develop a formula starting explicitly with those basic principles for how far on the rough surface m2

travel will before coming to rest if the kinetic friction coefficient is ?. Simplify your result as much as

possible.

Explanation / Answer

let u1 is the speed of m1 at the bottom

1) 0.5*m1*u1^2 = m1*g*h

==> u1 = sqrt(2*g*h)

here, m2 = 2*m1, u2 = 0

after the collision the velocity of second block,

v2 = 2*m1*u1/(m1+m2)

= 2*m1*sqrt(2*g*h)/(3*m1)

= (2/3)*sqrt(2*g*h)

let mue is the coefficient of the fiction on rough surfac

let d is the maximum distance travelled by m2 before coming to stop.

0.5*m2*v2^2 = mue*m2*g*d

0.5*2*g*h = mue*g*d

==> d = h/(2*mue)

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.