Problem 1 400 ms A bullet m 6g moving with an initial speed

Problem 1. 400 m/s A bullet (m- 6g) moving with an initial speed of 400 m/s is fired into a block (m = 1.1 kg), initially at rest, and passes through, as shown in the figure. The block is connected to a spring with force constant 815 N/m. If the block moves 7.0 cm to the right after impact, find: 5.00 cm- a) the speed of the bullet upon emerging from the block b) the mechanical energy converted into internal energy in the collision

Solution

here,

mass of bullet , m1 = 0.006 kg

mass of block , m2 = 1.1 kg

initial speed of bullet , u1 = 400 m/s

let the final speed of bullet be v1 and the box be v2

for spring

0.5 * k * x^2 = 0.5 * m2 * v2^2

815 * 0.07^2 = 1.1 * v2^2

v2 = 1.9 m/s

using consrvation of momentum

m1 * u1 = m1 * v1 + m2 * v2

0.006 * 400 = 0.006 * v1 + 1.1 * 1.9

v1 = 50.7 m/s

the final speed of bullet is 50.7 m/s

b)

the mechanical energy conseverted into internal energy , ME = 0.5 * m * u1^2 - 0.5 * m1 * v1^2 - 0.5 * k * x^2

ME = 0.5 * 0.006 * 400^2 - 0.5 * 0.006 * 50.7^2 - 0.5 * 815 * 0.07^2

ME =    470.3 J