Billiard ball A of mass mA = 0.120kg moving with speed of va= 2.8m/s strikes bal

Question

Billiard ball A of mass mA = 0.120kg moving with speed of va= 2.8m/s strikes ball B, initially at rest, of mass mb = 0.140kg. As a result, of the collision, ball A is deflected off at an angle of 30 degrees with a speed v'a = 2.1m/s.  (a) Taking the x-axis to be the original direction of motion of ball A, write down the equations expressing the conservation of momentum for the for the components in the x and y direction separately. (b) Solve these equations for the speed of v'b and the angle (theta)b of the ball C after the collision. Do not assume the collision is elastic.

Explanation / Answer

a)

in x direction

0.12*2.8 = 0.12*2.1*cos(30) + 0.14*x

y direction

0 =0.12*2.1*sin(30) + 0.14*y

b) x = 0.841

y = -0.9

so vb' = sqrt(0.841^2 + 0.9^2)= 1.23

angle = arctan( 0.9/0.841)= - 46.94 degrees

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