(1) Two curling rocks of equal mass, one orange and the other yellow, are involv

Question

(1) Two curling rocks of equal mass, one orange and the other yellow, are involved in a perfectly elastic, glancing collision. The yellow rock is initially at rest and is struck by the orange rock which is moving at a speed voi “ 6 m/s. After the collision, the orange rock moves along a direction that makes an angle of “ 20o with its initial direction of motion. (a) What is the magnitude of the velocity of the orange rock after the collision? (b) What is the velocity of the yellow rock after the collision? (c) Using your results for parts (a) and (b), find the angle between the velocity of the yellow rock and the initial direction of motion.

Explanation / Answer

initial momentum of the system is pi = m*voi = 6*m

final momentum along original direction is pf = (m*vof*cos(20))+(m*vyf*cos(alpha))

now using law of conservation of momentum

6*m = (m*vof*cos(20))+(m*vyf*cos(alpha))

m cancels

6 = 0.94*vof + vyf*cos(alpha)

vyf*cos(alpha) = 6-(0.94*vof).........(1)

momentum along perpendicular to the original direction

0 = (m*vof*sin(20))-(m*vyf*sin(alpha))

0.342*vof - (vyf*sin(alpha)) = 0

vyf*sin(alpha) = 0.342*vof ........(2)

and also using law of conservation of kinetic energy

0.5*m*6^2 = (0.5*m*vof^2)+(0.5*m*vyf^2)

18 = (0.5*vof^2)+(0.5*vyf^2).....(3)

(1)^2 + (2)^2

vyf^2 = (6-(0.94*vof))^2 + (0.342*vof)^2

substitute in (3) we get

18 = (0.5*vof^2)+(0.5*(6-(0.94*vof))^2 +(0.342*vof)^2)

vof = 5.32 m/sec is the answer for a)

from (3)

18 = (0.5*5.32^2)+(0.5*vyf^2)

vyf = 2.77 m/sec is the answer for b)

vyf*sin(alpha) = 0.342*vof ........(2)

2.77*sin(alpha) = 0.342*5.32

sin(alpha) = (0.342*5.32)/2.77

sin(alpha) = 0.656

alpha = sin^(-1)(0.656) = 41 degrees below the initial direction of motion is the answer for C)

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