Question

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. 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.

Answer 1

Conservation of momentum for a two-dimensional collision requires setting initial and final momenta equal in each direction separately. The equations take into account the trigonometric resolution of ball A's velocity post-collision and enable solving for the final velocity components of ball B.

Step-by-step explanation:

The question involves applying the conservation of momentum in a two-dimensional collision. In the x-direction, the conservation of momentum before and after the collision can be expressed as:

• Initial Momentum (mA × vA) + (mB × vB) = Final Momentum (mA × v'A,x) + (mB × v'B,x),
• 0.120 kg × 2.8 m/s + 0 = 0.120 kg × 2.1 m/s × cos(30°) + 0.140 kg × v'B,x

For the y-direction:

• mA × vA,y + mB × vB,y = mA × v'A,y + mB × v'B,y,
• 0 + 0 = 0.120 kg × 2.1 m/s × sin(30°) + 0.140 kg × v'B,y

By solving these equations, you can find the components of the final velocity of ball B. Notice the use of trigonometric functions to resolve the velocity of ball A after the collision into x and y components.