A 2.00 kg block is moving on a horizontal surface with velocity v_0 = 14.0 m/s w

Question

A 2.00 kg block is moving on a horizontal surface with velocity v_0 = 14.0 m/s when it has a perfectly elastic collision with a stationary block with a mass of 3.00 kg. See figure below. After the collision, the 3.00 kg block travels a distance d to point A, where it has speed v and begins projectile motion. The 3.00 kg block then impacts the ground at point B, a horizontal distance d from where it began projectile motion. This d is the same horizontal distance d where there was friction. Find v and d. Ignore friction except where noted.

Explanation / Answer

As given in the diagram,

consider collision of the two boxes
let velocity of box on the right be u after collision
then form conservation of momentum

2*14 = 2*v + 3*u
u = (28 - 2v)/3

also
then from conservation of energy of a perfectly elastic collision

0.5*2*14^2 = 0.5*3*u^2+ 0.*2*v^2 = 1.5(28 - 2v)^2 + v^2 = 1.5(784 + 4v^2 - 112v) + v^2 = 1176 + 7v^2 - 168v
7v^2 - 168v + 980= 0
v = 10 m/s
u = 2.667 m/s

KE of block aftger collision = 0.5*3*2.667^2 = 10.66933 J
energy lost in frivtion = mu*mg*d = 0.5*3*9.81*D = 14.715d
net energy at the end of track, 10.66933 - 14.715d = 0.5*3*v^2

and the horizontal distance travelled be d'
d' = vt
where 1.25 = 0.5*9.81t^2
d' = sqroot((10.66933 - 14.715d)/(0.5*3))*sqroot(1.25/0.5*9.81) = 0.4121sqroot((10.66933 - 14.715d))
but d' = d
so,
d^2 = 0.16989(10.66933 - 14.715d)
d^2 + 2.5d - 1.8126 = 0

d = 0.5871m
v = sqroot((10.66933 - 14.715d)/(0.5*3))1.1633 m/s =

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