We observe a glancing collision between two billiard balls of the same mass. The

Question

We observe a glancing collision between two billiard balls of the same mass. The first ball is incident at a speed of 5.54 m/s, strikes the second ball (initially at rest) and moves off with a speed of 4.62 m/s at an angle of 33.5° counterclockwise from the original line of motion. The second ball is initially at rest and after the collision moves off with a velocity which we wish to describe with respect to the first ball's original line of motion. Determine the following.

(a) Ignore the fact that the two balls are rotating and determine the velocity (magnitude and direction) of the second ball after the collision.

Explanation / Answer

here we use the conservation of momentum

first in x direction , and second ball initial velocity = 0 because it starts from rest

Pxi = Pxf

m(5.54) + m*0 = m(4.62) cos33.5 + m(v2f)cosphi

v2fcosphi = 5.54-(4.62*cos33.5)

v2fcosphi = 1.69 m/s ...(1)

momentum conservation in y direction

intially there is no componet so is zero

Piy = Pfy

0 + 0 = m*4.62*sin33.5 + mv2fsinphi

v2sinphi = -2.55 m/s ...(2)

squaring equation [1 and 2 ] and add them

v^2(cos^2phi + sin^2phi) = (1.69)^2 + (-2.55)^2

and cos^2phi + sin^2phi = 1

v^2 = (1.69)^2 + (-2.55)^2

v = sqrt((1.69)^2 + (-2.55)^2 )

v = 3.059 m/s = 3.06 m/s

b) direction

divide equation 2 by equation 1

sinphi/cosphi = -2.55/1.69

tanphi = -2.55/1.69

phi = tan^-1(-2.55/1.69)

phi = -56.47 degree below the x axis

Related Questions

Navigate

• Browse (All)
• Browse W
• Subjects

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