Question

0.150 kg stone rests on a frictionless, horizontal surface. A bullet of mass 7.00 g , traveling horizontally at 300 m/s , strikes the stone and rebounds horizontally at right angles to its original direction with a speed of 240 m/s .

A)Compute the magnitude of the velocity of the stone after it is struck.

B)Compute the direction of the velocity of the stone after it is struck.

C)Is the collision perfectly elastic?

Answer 1

here,

the mass of bullet ,m = 7 g = 0.007 kg

mass of stone , M = 0.15 kg

initial speed of bullet , u1 = 300 m/s i

the final speed of bullet , v1 = 240 m/s j

let the final speed of stone be v2

using conservation of momentum

m * u1 = m * v1 + M * v2

0.007 * 300 i = 0.007 * 240 j + 0.15 * v2

v2 = (14 i m/s - 11.2 j m/s)

a)

the magnitude of velocity , |v2| = sqrt(14^2 + 11.2^2) = 17.9 m/s

b)

the direction of velocity , theta = arctan(11.2/14) = 38.7 degree from the inital direction of bullet

c)

the initial kinetic energy , KEi = 0.5 * m * u^2

KEi = 0.5 * 0.007 * 300^2 = 315 J

the final kinetic energy , KEf = 0.5 * m * v1^2 + 0.5 * M * v2^2

KEf = 0.5 * 0.007 * 240^2 + 0.5 * 0.15 * 17.9^2 = 225 J

as KEf is not equal to KEi

the collison is NOT perfectly elastic