Speed and Direction after Collision

[jeeves_17]

A 20 g ball of clay traveling east at 2.0 m/s collides with a 30 g ball of clay traveling 300 south of west at 1.0 m/s . What are the speed and direction of the resulting 50g blob of clay?



Relevant equations
Pi = Pf



The attempt at a solution
(Px)i = mF(-Vfcostheta) + mP (Vp)
= (0.03 kg)(-1.0)(cos30) + (0.02kg)(2.0)
= 0.014019

(Py)i = mf(Vfy)i + mP(Vpy)i = mf(-vfsintheta) + 0
= (0.3)(1.0)(sin30) = 0.15 kg.m/s

(Px)f = (Px)i
(Vx)f = (Px)i / (mF + mP) = 0.014019 / (0.03 +0.02)
= 0.28038 m/s
(Py)f = (Py)i
(Vy)f = (Py)i / (mF + mP) = 0.15 / (0.02 +0.03) = 3.0 m/s

Direction = tan-1 [(Vy)f / (Vx)f] = tan-1 [ 3 / 0.28038] = 84.66 Degrees

V = SquareRoot[ (Vx)f^2 + (Vy)f^2) = Squareroot[ 3^2 + 0.28038^2] = 3.013 m/s



IS this Right??

 

[Bill Foster]

A 20 g ball of clay traveling east at 2.0 m/s collides with a 30 g ball of clay traveling 300 south of west at 1.0 m/s . What are the speed and direction of the resulting 50g blob of clay?

What does that mean "300 south of west"? 300 degrees?

Start with simple conservation of momentum:

$$m_1 v_1 + m_2 v_2 = (m_1 + m_2)v$$

Now in vector notation:

$$m_1 \vec v_1 + m_2 \vec v_2 = (m_1 + m_2) \vec v$$

Given the specifics of the problem, we can conclude that the motion is limited to two dimensions: east-west and north-south. We'll let the east-west be the x component and the north-south be the y component.

So:

$$\vec v_1 =v_1\cos{\theta_1}\hat x+v_1\sin{\theta_1}\hat y$$
$$\vec v_2 =v_2\cos{\theta_2}\hat x+v_2\sin{\theta_2}\hat y$$
$$\vec v =v\cos{\theta_3}\hat x+v\sin{\theta_3}\hat y$$

Now we can write the whole mess out...

$$m_1 (v_1\cos{\theta_1}\hat x+v_1\sin{\theta_1}\hat y) + m_2 (v_2\cos{\theta_2}\hat x+v_2\sin{\theta_2}\hat y) = (m_1 + m_2) (v\cos{\theta_3}\hat x+v\sin{\theta_3}\hat y)$$

Now we can group the individual vector components:

$$m_1 v_1\cos(\theta_1)\hat x + m_2 v_2\cos(\theta_2)\hat x = (m_1 + m_2) v\cos(\theta_3)\hat x$$
$$m_1 v_1\sin(\theta_1)\hat y + m_2 v_2\sin(\theta_2)\hat y = (m_1 + m_2) v\sin(\theta_3)\hat y$$

Now you have two equations and two unknowns.

Reduce:

$$\frac{m_1 v_1\cos(\theta_1) + m_2 v_2\cos(\theta_2)}{m_1 + m_2} = v\cos(\theta_3)$$

$$\frac{m_1 v_1\sin(\theta_1) + m_2 v_2\sin(\theta_2)}{m_1 + m_2} = v\sin(\theta_3)$$

Now it's easy to find the angle. Just divide the equations, resulting in this:

$$\frac{m_1 v_1\sin(\theta_1) + m_2 v_2\sin(\theta_2)}{m_1 v_1\cos(\theta_1) + m_2 v_2\cos(\theta_2)}=\tan{\theta_3}$$

To find $$v$$, square both equations and add them together...

$$v^2\cos^2{\theta_3}+v^2\sin^2{\theta_3}=...$$

Use the trig identity:

$$\cos^2{\theta}+\sin^2{\theta}=1$$

$$v^2(\cos^2{\theta_3}+\sin^2{\theta_3})=...$$

$$v^2=...$$

 

[jeeves_17]

sorry it's 30 degrees

 

[jeeves_17]

i got an angle of 52 degrees and a velocity of 0.94 .. is that right:S