Question

69) A particle with a mass of 400 g movig at 2.0 m/s collides elastically with a stationary particle whose mass is 240 g.

After the collision, the 400 g mass is moving at 1.5 m/s. Through what angle is its path deflected?

The answer is 30. Please show all the work with fomular.

Answer 1

let iniital velocity of 400 grams mass is along +ve x axis.

let i and j are unit vectors along +ve x and +ve y axis.

then total initial momentum=0.4*(2 i)+0.24*0=(0.8 i ) kg.m/s

let path of the 400g mas deflected by angle theta .

then its velocity in vector form=1.5*(cos(theta) i +sin(theta) j)

let speed of the sttionary particle after collision be v1.

as collision is elastic, conserving kinetic energy before and after collision:

0.5*0.4*2^2+0.5*0.24*0^2=0.5*0.4*1.5^2+0.5*0.24*v1^2

==>v1=1.7078 m/s

let angle of this velocity with +ve x axis is alpha.

then in vector notation, velocity =1.7078*(cos(alpha) i+sin(alpha) j)

writing conservtion of momentum:

0.8 i=0.4*1.5*(cos(theta) i+sin(theta) j)+0.24*1.7078*(cos(alpha) i+sin(alpha) j)

equating components along x and y axis:

0=0.4*1.5*sin(theta)+0.24*1.7078*sin(alpha)

==>sin(theta)=-0.68312*sin(alpha)...(1)

and 0.8=0.4*1.5*cos(theta)+0.24*1.7078*cos(alpha)

==>cos(theta)=1.33-0.68312*cos(alpha)..(2)

squaring and adding equations 1 and 2:

0.68312^2*sin^2(alpha)+1.33^2+0.68312^2*cos^2(alpha)-2*1.33*0.68312*cos(alpha)=sin^2(theta)+cos^2(theta)

==>0.68312^2+1.33^2-2*1.33*0.68312*cos(alpha)=1

==>cos(alpha)=0.67996

==>alpha=47.16 degrees

using this value of alpha in equation 1,

theta=-30 degrees

so angle is 30 degrees below the original direction of motion.