please solve A 0.30 kg puck, initially at rest on a frictionless horizontal surf
ID: 2210589 • Letter: P
Question
please solve
A 0.30 kg puck, initially at rest on a frictionless horizontal surface, is struck by a 0.20 kg puck that is initially moving along thexaxis with a velocity of2.8m/s. After the collision, the 0.20 kg puck has a speed of0.5m/s at an angle of?= 53
Explanation / Answer
detailed answer M1 = 0.30 kg and V1= 0 M2 = 0.20 kg and V2= 2.3 m/s V2' = 0.5 m/s (? = 53° to the positive x axis) => (V2')x = (V2') cos 53° = 0.3 m/s .....(in the direction of x axis x) => (V2')y = (V2') sin 53° = 0.4 m/s ......(in the direction of y axis) then, the speed vector : V2= 2.3 i + 0 j V2' = 0.3 i + 0.4 j ......(i and j are unit vectors) (a). Use formula : => M1V1+ M2V2 = M1(V1)' + M2(V2)' => 0 + 0.20 (2.3 i) = 0.30 (V1)' + 0.20 (0.3 i + 0.4 j) => 0.46 i = 0.30 (V1)' + (0.06 i + 0.08 j) => 0.46 i - (0.06 i + 0.08 j) = 0.30 (V1)' => 0.4 i - 0.08 j = 0.30 (V1)' => (V1)' = (0.40 i - 0.08 j) / 0.30 => (V1)' = 1.333 i - 0.267 j (m/s) .... (final speed vector of M1) =>¦(V1)'¦= v(1.333² + 0.267²) =>¦(V1)'¦= v(1.848) =>¦(V1)'¦= 1.36 m/s ....... (magnitude final speed of M1) the direction of (V1)' : use formula : tan ? = (Vy) / (Vx) => tan ? = (- 0.267) / (1.333) => tan ? = - 0.2003 => ? = - 11,326° .......(from +x axis) (b). The fraction of kinetic energy (KE) lost in the collision. ? total initial of kinetic energy : => S KE = KE1+ KE2 => S KE = (½ M1V1²) + (½ M2V2² ) => S KE = 0 + (½ x 0.30 x 2.3²) => S KE = 0.7935 Joule ? total final of kinetic energy : => S (KE)' = (KE1)' + (KE2)' => S (KE)' = { ½ M1((V1)')² } + { ½ M2((V2)')² } => S (KE)' = { ½ (0.30) (1.36)² } + { ½ (0.20) (0.5)² } => S (KE)' = 0.277 + 0.025 => S (KE)' = 0.302 Joule Then : ?KE = S KE - (S KE)' => ?KE = 0.7935 - 0.302 => ?KE = 0.4915 Joule
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