A bullet, B, weighing 162 gr (1 lb = 7,000 gr) is fired with speed v0 as shown a

Question

A bullet, B, weighing 162 gr (1 lb = 7,000 gr) is fired with speed v0 as shown and becomes embedded in the center of a rubber block of dimensions h = 4.15 in. and w = 6.19 in. weighing Wrb = 2.5 lb. The rubber block is attached to the end of a uniform thin rod, A, of length L = 1.50 ft, weight Wr = 5.1 lb, and that is pinned at O. After the impact, the rod (with the block and the bullet embedded in it) swings upward to an angle of 62degree. Determine the speed of the bullet right before impact. v0 = ft/s

Explanation / Answer

cos62=x/1.5 x=0.704ft vertical height the block rise due to impact of bullet=1.5-0.704=0.796ft k.e of bullet=p.e of block 1/2mv^2=mgh 1/2[(162/7000)]*v^2=2.5*32.2*0.796 v=74.4ft/s

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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