Puck A (with mass 0.250 kg) is travelling toward puck B (with mass 0.350 kg), which is originally at rest, on a frictionless, horizontal air table. Puck A has a velocity to the left of 0.120 m/s after the contact, while puck B has a velocity to the right of 0.650 m/s.(a) How fast was puck A traveling before to the collision?(b) Determine the change in the system's overall kinetic energy that results from the impact.

Question

Puck A (with mass 0.250 kg) is travelling toward puck B (with mass 0.350 kg), which is originally at rest, on a frictionless, horizontal air table. Puck A has a velocity to the left of 0.120 m/s after the contact, while puck B has a velocity to the right of 0.650 m/s.(a) How fast was puck A traveling before to the collision?(b) Determine the change in the system&#039;s overall kinetic energy that results from the impact.

pseudoenergy34 · Accepted Answer

The masses of pucks A and B are &amp;nbsp;mA=0.250kg&amp;nbsp;and&amp;nbsp;mB=0.350kg&amp;nbsp;respectively.Using right as the constructive axis, we have:v→A1=???&amp;nbsp;v→A2=−0.120m/s&amp;nbsp;v→B1=0&amp;nbsp;v→B2=0.650m/sWe are aware that a system&#039;s overall momentum is the vector sum of the moments of each of the system&#039;s constituent particles:&amp;nbsp;p→=p→A+p→B+&amp;nbsp;... (1)Where a particel&#039;s momenta are determined by:&amp;nbsp;p→=mv→&amp;nbsp;(2)Additionally, we know that a system&#039;s momentum is conserved for an isolated system, hence&amp;nbsp;p→=&amp;nbsp;constant.a) The system (pucks A and B) is regarded as an isolated system because the table is frictionless. In accordance with the principle of momentum conservation, the system&#039;s momentum before and after the collision are equal, leading to the following results:&amp;nbsp;p→1=2→2&amp;nbsp;(3)Applying equation (1) and (3), we get:&amp;nbsp;p→A1

On a frictionless, horizontal air table, puck A (with mass 0.250 kg) is moving t

Answered question

Answer & Explanation

New Questions in Differential Calculus