I need help with number 2 Collisions 2 m1 m2 A block with a mass of m1- 5.3 kg i

Question

I need help with number 2

Collisions 2 m1 m2 A block with a mass of m1- 5.3 kg is pushed to the right with a force of 769 N for a distance of 3.2 m across a horizontal frictionless surface, after which the force is removed. The 5.3-kg block then collides with a second block with a mass m2 -4.6 kg, which is initially at rest. The two blocks stick together after the collision. 1 Determine the change in the internal energy of the system of blocks during the collision. 1145.15 Submit Your submissions: 1145.15 Computed value: 1145.15 Feedback: Your answer has been judged correct; the exact answer is: 1143.40202020202 Submitted: Wednesday, April 18 at 10:30 AM 2) Now assume that the 4.6-kg block is traveling at 5.2 m/s to the left before the collision. The two blocks will still stick together after the collision. Determine the change in the internal energy of the system of blocks. V1 V2 m1 m2

Explanation / Answer

1) Work done by force = change in KE

(769 x 3.2) = (5.3) v0^2 /2

v0 = 30.473 m/s

Now applying momentum conservation for collision,

(5.3 x 30.473) + (4.6 x 0) = (4.6 + 5.3) v

v = 16.314 m/s

ki = 5.3 v0^2 /2 = 2460.8 J  

Kf = (5.3 + 4.6) v^2 /2 =1317.398 J  

change in thermal enery = Ki - Kf = 1143.4 J  

2) pi = pf

(5.3 x 30.473) + (4.6 x - 5.2) = (5.3 + 4.6) v

v = 13.898 m/s

ki = 5.3v0^2 /2 + 4.6(5.2^2)/2 = 2522.992 J

Kf = (5.3 + 4.6) v^2 / 2 = 956.068 J

thermal energy = 1567 J

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---

---

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