An 8.00 kg mass moving east at 16.0 m/s on a frictionless horizontal surface col

Question

An 8.00 kg mass moving east at 16.0 m/s on a frictionless horizontal surface collides with a 10.0 kg object that is initially at rest. After the collision, the 8.00 kg object moves south at 4.30 m/s. (a) What is the velocity of the 10.0 kg object after the collision? m/s (magnitude)

Explanation / Answer

(a) By conserving momentum in two perpendicular directions we get 8*16.0 + 0 = 0 + 10*Vx => Vx =12.84 m/s and 0 + 0 = 8*4.3 -10*Vy => Vy = 3.44 m/s magnitude = sqre root{(11.84*11.84) + (3.44*3.44)} = 12.3296 m/s=12.33 m/s (b) initial KE = 1/2 *8*14.8*14.8=876.16 J final KE = 1/2*8*4.3*4.3 + 1/2 *10*12.33*12.33 = 73.96 + 760.15 = 834.11 J hence percentage loss in KE = (dK/initial KE)*100 = 4.79 % KE is lost !

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.