A small ball A of mass m_A = 0.120 kg moving with speed v_A = 2.80 m/s strikes b

Question

A small ball A of mass m_A = 0.120 kg moving with speed v_A = 2.80 m/s strikes ball B, initially at rest, of mass m_B = 0.140 kg. As a result of the collision, ball A is deflected off at an angle of 30.0 degree with a speed v_B = 2.10 m/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 components in the x and y directions separately. (b) Solve these equations for the speed v_B and angle, theta_B of ball B. Do not assume the collision is elastic.

Explanation / Answer

(a).

P = 0

ma va - mava' - mb vb' = 0

0.12 (2.8){1,0} - 0.12 (2.1)(cos 30, sin 30) - vb' (0.14) (cos , sin ) = 0



(b).

0.12 (2.8){1,0} - 0.12 (2.1)(cos 30, sin 30) - vb' (0.14) (cos , sin ) = 0

0.12 (2.8) - 0.12 (2.1) cos 30 = vb' (0.14) cos ................ (1)

0 - 0.12 (2.1) sin 30 = vB' (0.14) sin ...............................( 2)

equation 2 divided by equation 1,

tan = (0 - 0.12 (2.1) sin 30°)/(0.12 (2.8) - 0.12 (2.1) cos 30°)

= -46.94°

back to equation 2,

0 - 0.12 (2.1) sin 30° = vb' (0.14) sin (-46.94°)

vb' = 1.2318 m/s

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