please show your work Two blocks are free to slide along the frictionless wooden

Question

please show your work

Two blocks are free to slide along the frictionless wooden track shown below. The block of mass m1 4.99 kg is released from the position shown, at height h 5.00 m above the flat part of the track. Protruding from its front end is the north pole of a strong magnet, which repels the north pole of an identical magnet embedded in the back end of the block of mass m2 9.90 kg, initially at rest. The two blocks never touch, but we can treat this as a perfectly elastic collision. Calculate the maximum height to which m1 rises after the elastic collision. rt m2

Explanation / Answer

By Conservation of energy we find the speed of block m1 at the bottom of the track

(1/2)m1V12 =m1gh

(1/2)*V12 =9.81*5

V1=9.905 m/s

Final Speed of Block m1 after the elastic collision is

V1f =(m1-m2)V1/(m1+m2) + 2m2V2/(m1+m2)

V1f =(4.99-9.9)*9.905/(4.99+9.9) + 0

V1f=-3.266 m/s

After collision ,by Conservation of energy

(1/2)mV1f2 =mghmax

(1/2)*(-3.266)2 =9.81*hmax

hmax=0.54368 m

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---

---

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