A 7 g bullet is travelling at a speed of 120 m/s towards a 800 g block which lie

Question

A 7 g bullet is travelling at a speed of 120 m/s towards a 800 g block which lies on a horizontal and frictionless surface at the end of a string of length 0.7 m. The bullet's trajectory is tangent to the circle defined by the string. (Figure 1) 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 800 g block after the bullet exits? Got 0.525 m/s Correct
B. After the collision, the 800 g block moves in a circle defined by the string. What is the tension in the string? Not sure

Explanation / Answer

a) initial momentum = final momentum 7*120 = 60 *7 + 800 * x x = 0.525 m/s mv^2/r = t = 0.315 N Not yet rated Emily Granger - 6 minutes later This collision is inelastic. In this type of collision, momentum is conserved but kinetic energy is not. From the law of conservation of momentum (linear) : m1v1(i) +m2v2(i) = m1v1(f) + m2v2(f) Since the block is initially at rest, m2v2(i) = 0. Therefore : m1v1(i) = m1v1(f) + m2v2(f) Solving for v1(f) : v1(f) = [m1v1(i) - m2v2(f)] / m1 = [(0.0042kg)(650m/s) – (0.096kg)(21m/s)] / 0.0042kg = 170m/s The kinetic energy of the system? Not sure what is meant by that. The KE of the bullet before the collision is : KE = 0.5mv² = 0.5(0.0042kg)(650m/s)² = 890J (rounded).

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