4) Box A, of mass m, slides without friction over the icy hill of height h=8.50

Question

4)

Box A, of mass m, slides without friction over the icy hill of height h=8.50 m, shown in the

figure. Once at the top of the hill, it experiences an elastic collision with a box B, of mass 2m.

Box B lands 6.20 m from the foot of the cliff, with no air resistance. Box A slides back down the

hill after the collision.

Determine box As initial speed v0 at the bottom of the hill. Also determine the final speed of

box A after it slides back down the hill (after the collision) to this same position. (Hint: there are

many different moments in time you need to considering here; keep track of your subscripts!)

The answers are

vfinal = 13.1 m/s

v0 = 14.7 m/s

I would appreciate to know how one solves this problem. Thank you.

Explanation / Answer

let t is the time taken to land the box B,

h = 0.5*g*t^2

t = sqrt(2*h/g)

= sqrt(2*8.5/9.8)

= 1.317 s

velocity of box B after the collision,
Vbf = x/t = 6.2/1.317 = 4.707 m/s

let vai is the velocity of box A before hitting box B.
we know, Vbf = 2*ma*vai/(ma+mb)

vbf = (2/3)*vai

==> vai = (3/2)*4.707 = 7.061 m/s

let uai is the initial velocity of box A at the bottom.

vai^2 - uai^2 = -2*g*h

uai = sqrt(vai^2 + 2*g*h)

= sqrt(7.061^2 + 3*9.8*8.5)

= 14.71 m/s (answer for vo) <<<<<<<<--------Answer


let vaf is the velocity box A after the collsion

vaf = (ma-mb)*vai/(ma+mb)

= (-1/3)*7.061

= -2.354 m/s (in th opposite direction)

vaf = 2.354 m/s (speed)(magnitude)

let Vf is the final speed of box A at the bottom at the bottom.

Vf^2 - vaf^2 = 2*g*h

Vf = sqrt(vaf^2 +2*g*h)

= sqrt(2.354^2 + 2*9.8*8.5)

= 13.12 m/s <<<<<<<<<<<------------Answer

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