(a) 0.5 kg m/s
Before the collision, only the first body is moving, so only the first body contributes to the total momentum.
The first body has
m = 400 g = 0.4 kg (mass)
v = 1.25 m/s (velocity, towards east direction)
So its momentum is
[tex]p_x=mv = (0.4 kg)(1.25 m/s)=0.5 kg m/s[/tex]
And since the body is moving along the east direction, this is also the easterly component of the total momentum before the collision.
(b) Zero
Before the collision, we have:
- The first body moving along the east direction --> so its northerly component is zero
- The second body at rest --> this means that it does not contribute to the momentum, since it is zero
This means that the northerly component of the total momentum before the collision is zero.
(c) 0.5 m/s at 53.1 degrees south of east
The law of conservation of momentum states that each component of the total momentum must be conserved.
- Along the easterly direction:
[tex]p_x = p_{1x} + p_{2x}[/tex]
where
[tex]p_x = 0.5 kg m/s[/tex] is the easterly component of the total momentum
[tex]p_{1x} = m v cos \theta = (0.4 kg)(1.00 m/s) cos 36.9^{\circ} =0.32 kg m/s[/tex] is the easterly component of the momentum of the first body after the collision
[tex]p_{2x}[/tex] is the easterly component of the momentum of the second body (mass m = 600 g = 0.6 kg) after the collision
Solving the equation we find
[tex]p_{2x} = p_x - p_{1x} = 0.5 kg m/s - 0.32 kg m/s = 0.18 kg m/s[/tex]
- Along the northerly direction:
[tex]p_y = p_{1y} + p_{2y}[/tex]
where
[tex]p_y = 0 kg m/s[/tex] is the northerly component of the total momentum
[tex]p_{1y} = m v sin \theta = (0.4 kg)(1.00 m/s) sin 36.9^{\circ} =0.24 kg m/s[/tex] is the northerly component of the momentum of the first body after the collision
[tex]p_{2y}[/tex] is the northerly component of the momentum of the second body (mass m = 600 g = 0.6 kg) after the collision
Solving the equation we find
[tex]p_{2y} = p_y - p_{1y} = 0 - 0.24 kg m/s = -0.24 kg m/s[/tex]
So now we find the momentum of the 600 g body after the collision:
[tex]p_2=\sqrt{p_{2x}^2 + p_{2y}^2}=\sqrt{(0.18)^2+(-0.24)^2}=0.3 kg m/s[/tex]
and so its final speed is
[tex]v=\frac{p_2}{m}=\frac{0.3 kg m/s}{0.6 kg}=0.5 m/s[/tex]
and the direction is
[tex]\theta=tan^{-1} (\frac{p_{2y}}{p_{2x}})=tan^{-1} (\frac{-0.24}{0.18})=-53.1^{\circ}[/tex]
so 53.1 degrees in the south-east direction.
