Question

Three balls of clay with m1=40 g, m2=30 g and m3=20 g are moving with velocities v1=4.0 m/s at 315 deg, v2=3.0m/s at 180 deg and v3=2m/s at 90deg respectively (all angles are given relative to the +x axis). The masses collide and create a single blob of clay. A)Determine the magnitude and direction of the blob after the collision. B)Determine the velocity of the center of mass before the collision.

Answer 1

A) By the conservation of linear momentum
Initial linear momentum=final linear momentum
(m1V1cos45)i^+(m1V1sin45)j^+(-m2V2)i^+(m3V3)j^=(m1+m2+m3)V
(m1V1cos45-m2V2)i^+(m1V1sin45+m3V3)j^=(m1+m2+m3)V
Hence,V={(m1V1cos45-m2V2)i^+(m1V1sin45+m3V3)j^}/(m1+m2+m3)
find magnitude of V from vector formula.
and direction \[\phi =\tan^{-1} (m1V1\sin45+m3V3)/(m1V1\cos45-m2V2) \]... find phi from here this gives you direction of V vector ,here phi is the angle between V and +ve Xaxis.
B)velocity of centre of mass at initial=Velocity of centre of mas at final=V
since ,there is no external force .so apply above formula.