A 5. 2 kg block with a speed of 15 m/s collides with a 15 kg block that has a sp

Question

A 5. 2 kg block with a speed of 15 m/s collides with a 15 kg block that has a speed of 6. 5 m/s in the same direction. After the collision, the 15 kg block is observed to be traveling in the original direction with a speed of 7. 6 m/s. (a) What is the velocity of the 5. 2 kg block immediately after the collision? (b) By how much does the total kinetic energy of the system of two blocks change because of the collision? (c) Suppose, instead, that the 15 kg block ends up with a speed of 9. 3 m/s. What then is the change in the total kinetic energy?

Explanation / Answer

here,

M1 = 5.2 kg
V1 = 15 m/s

M2 = 15 kg
V2 = 6.5 m/s
V2'= 7.6 m/s

PART A:)
From conservation of linear momentum we have
M1V1 + M2V2 = M1V1' + M2V2'
5.2 * 15 + 15 * 6.5 = 5.2 * V1' + 15 * 7.6
78 + 97.5 - 114 = 5.2V1'
V1' = -11.82 m/s

The velocity of block 1 just after collison is -11.82 m/s that is opposite to it's original direction .

PART B:)

KEi = 0.5 * M1 * V1 = 0. 5 * 5.2 * 15^2
KEi = 585 J

KEf = 0.5 * M1 *V1'^2 + 0.5 * M2 *V2'^2
KEf = 0.5 * 5.2 * (-11.82)^2 + 0.5 * 15 * 7.6^2
KEf = 796.45 J

Lost Kinectic Energy will be Equal to : 796.45 - 585 = 211.45

PARt C:
M2 = 15 kg
V2' = 9.3 m/s

KEf = 0.5 * M1 *V1'^2 + 0.5 * M2 *V2'^2
KEf = 0.5 * 5.2 * (-11.82)^2 + 0.5 * 15 * 9.3^2
KEf = 1011.92 J

Lost Kinectic Energy will be Equal to : 1011.92 - 585 = 800.47 J

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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