# Homework Help: What is the velocity after the collision?

1. Oct 4, 2004

### punjabi_monster

A 15.0 kg object is moving east at a velocity of 7.0 m/s when it collides with a 10.0 kg statinary object. After the collision the 15.0 object is travelling at a velocity of 4.2 m/s 20.0° S of E.

What is teh velocity of the 10.0 kg object after the collsion.

so far this is what i did, can you please tell me what i am doing worng? Thank-you.

15 kg object → a y→y component
10 kg object → b x→x component

Pa + Pb = P'a + P'b

Before Collision
Pa=mv
Pa=(15.0 kg)(7.0 m/s E)
Pa=105 Kgm/s

After Collsion
P'a=mv
P'a=(15.0 Kg)(4.2 m/s 20° S of E)
P'a=63 Kgm/s

P'b=mv
P'b=(10.0 Kg)(v)
P'b=?

Sin20 = P'ay/63 kgm/s
P'ay=21.5 kgm/s

Cos20°=P'ax / 63 kgm/s
P'ax = 59.2 kgm/s

Px=Squareroot[(105 kgm/s)^2 - (21.5 kgm/s)^2]
Px=102.8 kgm/s

P'bx=102.8 kg m/s - 59.2 kgm/s
P'bx= 43.6 kgm/s

P'b= squareroot[(43.6 kgm/s)^2 + (21.5 kgm/s)^2]
P'b= 48.6 kgm/s

P'b=mv
v=P'b/m
v=(48.6 kgm/s) / (10.0kg)
v= 4.86 m/s

CosX=(43.6 kgm/s)/(48.6 kgm/s)
X= 26° N of E

2. Oct 4, 2004

### HallsofIvy

You are writing this in a very strange way: I would not, for example, say that "Pa=105 Kgm/s" since Pa is a vector quantity. Other than that, I get very close to what you do- perhaps a roundoff error problem.

The way I would do this is write out the momentum vectors:
before the collision they are
<15*7, 0>= < 105, 0> and <0, 0> which adds, or course, to <105, 0>

After the collision, the momentum vector of the 15 kg mass is
<15*4.2*cos(20), 15*4.2*sin(-20)>= <59.2, -21.55>
We must have <px, py>+ <59.20, -21.55>= <105, 0> so
px= 105- 59.20= 45.80 and py= 0+ 12.55= 21.55
Since the momentum vector of this 10 kg mass is <45.80, 21.55>, its velocity vector is <4.580, 2.155>. If you want that in terms of speed and angle, the speed is
$\sqrt{4.580^2+ 2.155^2}= 5.06$ m/s and the angle is $tan^{-1}\frac{2.155}{4.58}= 25.2$ degrees north of east.

3. Oct 4, 2004

### punjabi_monster

thanks...i figured out my error. I should have used 105 Kgm/s for Px.