(3 marks) A 4.17 kg bowling ball (Ball B) collides with a 3.63 kg bowling ball (Ball A). Initially, Ball A is at rest while Ball B moves in the direction of the positive y-axis with a speed of 25

m/s. After the two balls collide, Ball B travels at 20 m/s at an angle 4π /3 anti-clockwise from the positive x-axis. Determine the speed (to the nearest m/s) and direction (to the nearest degree) of Ball A after the collision.

Community Answer

【General guidance】The answer provided below has been developed in a clear step by step manner.Step1/2Solution: Using conservation of momentum, we know that the total momentum before the collision is equal to the total momentum after the collision. Therefore:\( \mathrm{{m}_{{B}}\times{v}_{{B}}={m}_{{A}}\times{v}_{{A}}+{m}_{{B}}\times{v}_{{B}}'} \)where:\( \mathrm{{m}_{{B}}{\left(={4.17}\right)}} \) is the mass of Ball B\( \mathrm{{v}_{{B}}{\left(={25}\frac{{m}}{{\sec}}\right)}} \)is the velocity of Ball B before the collision\( \mathrm{{m}_{{A}}{\left(={3.63}{k}{g}\right)}} \) is the mass of Ball Av_A is the velocity of Ball A after the collision\( \mathrm{{v}_{{B}}'{\left(={20}\right)}} \)is the velocity of Ball B after the collisionPlugging in the given values, \( \mathrm{{4.17}\times{25}={3.63}\times{v}_{{A}}+{4.17}\times{20}} \)\( \mathrm{{v}_{{A}}={5.7438}} ... See the full answer