A 2500 kg car slides down a slippery road inclined at 15 degree. where the force

Question

A 2500 kg car slides down a slippery road inclined at 15 degree. where the force of friction is 6200 N. The car starts from rest and sides a distance of 6.5 m along the incline before hitting another car. What kinetic energy does the sliding car have upon impact? 1200 J 1760 J 1480 J 1320 J A proton traveling with speed 20 km's collides head on with a stationary positive ion. whose mass is six times that of the photon Assuming the collision is elastic, what is the final speed of the ion? 13 3 km/s 10.7 km/s 8.00 km/s 4.44 km/s A 10.0 g rifle bullet is fired with a speed of 600 m/s into a 5.00 kg wooden block that is initially at rest. The bullet remains lodged in the block. Compute the speed of the bullet and block together immediately after the bullet becomes lodged.

Explanation / Answer

8. Applying work -energy theorem,

Work done by all forces = change in KE

Work done by friction + work done by gravity = KEf - KEi

(6200 x 6.5 x cos180) + (2500 x 9.81 x 6.5 x sin15) = KE - 0

-40300 + 41259 = KE

KE = 959 J ...........Ans

9. suppose speed of electron is v1 (in opposite direction) and of ion is v2 after the collision.
for elastic collision,
velocity of approach = velocity of seperation
20 = v1 + v2

v1 = 20 - v2

now applying momentum conservation,

m (20) + 6m(0) = m(-v1) + 6m(v2)

20 = v2 - 20 + 6v2

v2 = 5.7 km/s .....Speed of ion

10. Applying momentum conservation on block-bullet system,

0.010 x 600 + 5 x 0   = (0.01 + 5)v

v = 1.198 m/s   Or 1.2 m/s

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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