Two brothers, Dustin and Parker, have a combined mass of 168 kg. Atan ice skatin
ID: 1671653 • Letter: T
Question
Two brothers, Dustin and Parker, have a combined mass of 168 kg. Atan ice skating rink, they stand close together on skates, at restand facing each other, with a compressed spring between them. thespring is kept from pushing them apart because they are holdingeach other. When they release their arms, Dustin moves off in onedirection at a speed of 0.90 m/s, while Parker moves off in theopposite direction at a speed of 1.2 m/s. Assuming that friction isnegligible, find Dustin’s mass.a. 72 kg
b. 80 kg
c. 96 kg
d. 77 kg
e. 84 kg
Explanation / Answer
According to law of conservation of linearmomentum (m1 + m2)* u = m1 *v1 + m2 *v2 initialvelocity u = 0 m is mass and v is final velocity, subscript1 referes to Parker and 2 refers to Dustin. (m1 + m2)* 0 = m1 * (-1.2) + m2 *0.90 - 1.2* m + 0.90 *m2 = 0 ---------(1) ( - ve sign with 1.2 is due to the fact thatParker moves in direction opposite Dustin) Also m1 + m2 = 168 ---------(2) Multiplying equation (2) by 1.2 and adding result to equation (1) 1.2 *m1 - 1.2 *m1 + 1.2 *m2 + 0.90 *m2 = 168 Dustin'smass m2 = 168/ 2.1 = 80.0 kg b. 80 kg initialvelocity u = 0 m is mass and v is final velocity, subscript1 referes to Parker and 2 refers to Dustin. (m1 + m2)* 0 = m1 * (-1.2) + m2 *0.90 - 1.2* m + 0.90 *m2 = 0 ---------(1) ( - ve sign with 1.2 is due to the fact thatParker moves in direction opposite Dustin) Also m1 + m2 = 168 ---------(2) Multiplying equation (2) by 1.2 and adding result to equation (1) 1.2 *m1 - 1.2 *m1 + 1.2 *m2 + 0.90 *m2 = 168 Dustin'smass m2 = 168/ 2.1 = 80.0 kg b. 80 kgRelated Questions
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