A particle of mass 8.83 × kg, moving at 1.48 × m/s, strikes an identical particl

Question

A particle of mass 8.83 × kg, moving at 1.48 × m/s, strikes an identical particle which is initially at rest. After the interaction, the particles (which can't be distinguished) are observed moving at angles 61.9° and 28.1°, both angles being measured with respect to the original direction of motion. What are the final speeds of the particles?
At 28.1° and 61.9°.

Explanation / Answer

m1 = m2 = m = 8.83 x kg u1 = 1.48 x m/s u2 = 0 ?1 = 61.9 degree ?2 = 28.1 degree v1 = ? v2 = ? Applying Conservation of momentum along the x-axis we have m1 * u1 = m1 * v1 cos?1 + m2 *v2 cos?2 ==> u1 = v1 * cos?1 + v2 *cos?2 ......1 Along the vertical direction we have v1 * sin?1 =v2 * sin?2 ........2 Solve eq 1 and 2 and substitute values , then you get the values of v1 and v2.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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