Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

As shown in the figure below, two blocks ( m 1 and m 2 ) are each released from

ID: 2259634 • Letter: A

Question

As shown in the figure below, two blocks (m1 and m2) are each released from rest at a height of

on a frictionless track and when they meet on the horizontal section of the track they undergo an elastic collision. If

and

determine the maximum heights to which they rise after the collision. Use the coordinate system shown in the figure.

y1f =   m y2f =   m. As shown in the figure below, two blocks (m1 and m2) are each released from rest at a height of h = 3.68 m on a frictionless track and when they meet on the horizontal section of the track they undergo an elastic collision. If m1 = 2.50 kg and m2 = 4.55 kg, determine the maximum heights to which they rise after the collision. Use the coordinate system shown in the figure.

Explanation / Answer

velocity of each point after reaching the bottom moost point.

V= sqrt (2*g*h) = srt(2*9.8*3.68) =8.493 m/s,

initial momentum, = 2.50 *8.493- 4.55*8.493 = -17.41

initial energy E = (2.5+4.55)*3.68*g = 254.51 J,

as collision is elastic, E and momentum must remain conserved,

let v1 and v2 be the velocity after colision,

final momentm= -17.41 = 2.55*vi1+4.55v2,

-17.41 = 2.55*v1+4.55v2

v1 = (-17.41-4.55v2) /2.55............(1)

final energy ,

254.51 = 1/2( 2.5v1^2 + 4.55 v2^2)..........

254.51= 1/2( 2.5{(-17.41-4.55v2) /2.55.}^2 + 4.55 v2^2)


solving these ,

v2 =1.91649 m/s,

v1 = -10.247 m/s

,

max height reached by the two blaocks,

h1 = v1^2/2g = 5.357 m

h2 = v2^2/2g = 0.1875m