Tilling in a table. put a box around each final answer you to work in groups of
ID: 1363658 • Letter: T
Question
Tilling in a table. put a box around each final answer you to work in groups of three or four and you may use your book, your own homework from this semester, and you notes. You may not use any other student's written work except in comparing answers. You should make sure YOU understand each step ( and that your are not simply copying your neighbor's work-that won't do you an good). TURN THIS IN AT THE END OF THE HOUR. A bullet of mass 0.120 kg travels with a speed of 55.0 m/s. It impacts a block of mass 0.750 kg which is attest on a table top as shown below. The coefficient of kinetic friction between the block and the table top is 0.175. Assume that the bullet imbeds itself in the block. Find the final velocity of the bullet block combination immediately after the collision, How much energy was lost in the collision? What impulse was delivered to the block during the collision? How far does the block side (use the work - energy theorem)?Explanation / Answer
here,
mass of block , M = 0.75 kg
speed of bullet , ub = 55 m/s
mass of bullet , m = 0.12 kg
(a)
let the final velocity of bullet and block be v
using conservation of momentum
m*ub = ( m + M) * v
0.12 * 55 = ( 0.12 + 0.75) * v
v = 7.59 m/s
the final velocity of the bullet and block is 7.59 m/s
(b)
energy lost in the collision , E = 0.5 * m * ub^2 - 0.5 * ( m + M)*v^2
E = 0.5 * 0.12 * 55^2 - 0.5 * 0.87 * 7.59^2
E = 156.44 J
the energy lost in the collision is 156.44 J
(c)
change in momentum for the block , P = Pf - Pi
P = 0.75 * 7.59 - 0
P = 5.7 kg.m/s
the impulse delivered to the block is 5.7 kg.m/s
(d)
uk = 0.175
accelration due to friction , a = uk*g
a = 1.72 m/s^2
v = 7.59 m/s
let the distance be d
v^2 - u^2 = 2*a*s
7.59^2 = 2 * 1.72 *d
d = 16.75 m
the distance d is 16.75 m