A 7 g bullet is travelling at a speed of 160 m/s towards a 600 g block which lie

Question

A 7 g bullet is travelling at a speed of 160 m/s towards a 600 g block which lies on a horizontal and frictionless surface at the end of a string of length .6 m. The bullet's trajectory is tangent to the circle defined by the string. The bullet goes completely through the block and exits with a speed of one-half its initial speed.
A)What is the velocity of the 600 g block after the bullet exits?
B)After the collision, the 600 g block moves in a circle defined by the string. What is the tension in the string?

Explanation / Answer

conservation of momentum gives m1u1=m1v1+m2v2 thus 7x160=7x80+600xv thus v=0.933m/s T-mg=mv^2/r thus T=(.6x9.8+.6x0.933^2/0.6=6.7449N

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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