Billiard ball A of mass mA = 0.120 kg moving with speed DA - 280 m/s strikes bal

Question

Billiard ball A of mass mA = 0.120 kg moving with speed DA - 280 m/s strikes ball B, at rest of mass mn = 0.140 kg. As a result of the collision, ball A is deflected off at an angle of with a speed D'A - 2.10 m/s. Taking the x axis to 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. solve these equations for the speed v's, and angle theta's of ball B. Do not assume the collision is classic.

Explanation / Answer

Applying conservation of momentum
along x axis

m1u1 + m2u2 = m1v1 +m2v2

here, m1 = 0.12 kg, m2 = 0.14 kg

u1 = 2.8 m/s, u2 = 0 m/s, v1 = 2.1 m/s, = angle with x axis

So, v2 = vx (velocity of ball 2 along x axis)

vy = velocity of ball 1 along y axis

0.12*2.8 + 0= 0.12 *2.1 cos+ 0.14 vx

along y axis
0 =0.12 *2.1sin +0.14 vy

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