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) I use principle of conservation of momentum (momentum is p=m.v mass times speed) in x and y direction.
for y :
m2 doesn't have momentum in y direction (angle is 180 degrees, so no velocity in y direction)
m3 has p3= m3.v.3 = 20g.-2m/s = -40 g.m/s
m1 has momentum p1=m1.v1.cos45=40g.4m/s.0.707=113.1g.m/s
I add this p1 and p3
py = p1 + p3 = -40 + 113.1 = 73,1 g m/s
now for x:
m3 has no velocity in x direction
so that leaves the two other masses m2 and m1
p1 in x direction is
p1 = m1 . v1y = 40g.-4m/s = -113,1 g.m/s (note here again in the negative x direction and with angle 315 degrees which is projected upon the x axis as cos 45 or cos 315)
p2 = m2.v2 = 30g.3m/s = 90g.m/s
px = p1x + p2x = -23,1 g.m/s

Now the total mass M = m1+m3+m3 = 90g

Now I apply the pricniple of conservation of momentum:
Px = M.Vx => Vx = px/M = -0.2566 m/s
Py = M.Vy => Vy = py/M = 0.8122m/s

So blob leaves towards second quadrant, angle is:
\[\theta=90^{0} + \tan^{-1} \left( 0.2566 \over0.8122 \right)\]
= 107,15 deg

Answer 2

sorry for the typos

Answer 3

speed is pythagoras rule for right triangle:
\[V = \sqrt{\left( vx ^{2} + vy ^{2} \right)}\]
= 0.85m/s

Answer 4

for b:
Determine the velocity of the center of mass before the collision.
=> is the same as after the collision, and that we allready calculated.
Center of mass size is 90 gram, it's velocity is 0.85m/s in direction 107,15 deg from x axis

Answer 5

Center of mass acts as if no collision happened.