A5.0-g particle moving 60 m/s collides with a 2.0-g particle initially at rest. after the collision, each of the particles has a velocity that is directed 30° from the original direction of motion of the 5.0-g particle. what is the speed of the 2.0-g particle after the collision?

after collision velocity is 87 m/s

Explanation:

given data

mass m1 = 5 g = 5× kg

velocity v1 = 60 m/s

mass m2 = 2 g

velocity v2 = 0 = 2× kg

angle 30°

to find out

after collision velocity v4

solution

we consider here v3 and v4 are velocity after collision

we know after collision 1st particle

v3 at x axis = v3cos30

v3 at y axis = v3sin30

and 2nd particle after collision

v4 at x axis = v4 cos 30

v4 at y axis = v4 sin30

we apply conservation of momentum here

so in x direction

m1v1 + m2v2 = m1v3 + m2v4     ...............1

put the value here

5× (60) + 0 = 5× ( v3cos30)  +  2× (  v4 cos 30 )

4.33 v3 + 1.732 v4 = 300    .......................2

in x direction

m1v1 + m2v2 = m1v3 + m2v4

5× (0) + 0  = 5× ( v3sin30)  -  2× ( - v4 sin 30 )

2.5 v3 - v4 = 0

v4 = 2.5 v3        ................................3

put equation 3 in equation 2

4.33 v3 + 1.732 (2.5 v3 ) = 300

v3 = 34.6

so

from equation 3

v4 = 2.5 (34.6)

v4 = 86.6

so after collision velocity is 87 m/s

it has to be b, about 97 percent.