Question

A particle with an initial linear momentum of 2.00 kg-m/s directed along the positive x-axis collides with a second particle, which has an initial linear momentum of4.00 kg-m/s, directed along the positive y-axis. The final momentum of the first particle is 3.00 kg-m/s, directed 45.0 above the positive x-axis.
a. the magnitude and direction (angle expressed counter-clockwise with respect to the positive x-axis) of the final momentum for the second particle
b. assuming that these particles have the same mass, % loss of their total kinetic energy after they collided

Answer 1

Answer:

a) p₂ = 1.88 kg*m/s

   θ = 273.4 º

b)  Kf = 37% of Ko

Explanation:

a)

• Assuming no external forces acting during the collision, total momentum must be conserved.
• Since momentum is a vector, their components (projected along two axes perpendicular each other, x- and y- in this case) must be conserved too.
• The initial momenta of both particles are directed one along the x-axis, and the other one along the y-axis.
• So for the particle moving along the positive x-axis, we can write the following equations for its initial momentum:

       $p_{o1x} = 2.00 kg*m/s (1)$

       $p_{o1y} = 0 (2)$

• We can do the same for the particle moving along the positive y-axis:

        $p_{o2x} = 0 (3)$

        $p_{o2y} = 4.00 kg*m/s (4)$

• Now, we know the value of magnitude of the final momentum p1, and the angle that makes with the positive x-axis.
• Applying the definition of cosine and sine of an angle, we can find the x- and y- components of the final momentum of the first particle, as follows:

       $p_{f1x} = 3.00 kg*m/s * cos 45 = 2.12 kg*m/s (5)$

      $p_{f1y} = 3.00 kg*m/s sin 45 = 2.12 kg*m/s (6)$

• Now, the total initial momentum, along these directions, must be equal to the total final momentum.
• We can write the equation for the x- axis as follows:

       $p_{o1x} + p_{o2x} = p_{f1x} + p_{f2x} (7)$

• We know from (3) that p₀₂ₓ = 0, and we have the values of p₀1ₓ from (1) and pf₁ₓ from (5) so we can solve (7) for pf₂ₓ, as follows:

       $p_{f2x} = p_{o1x} - p_{f1x} = 2.00kg*m*/s - 2.12 kg*m/s = -0.12 kg*m/s (8)$

• Now, we can repeat exactly the same process for the y- axis, as follows:

       $p_{o1y} + p_{o2y} = p_{f1y} + p_{f2y} (9)$

• We know from (2) that p₀1y = 0, and we have the values of p₀₂y from (4) and pf₁y from (6) so we can solve (9) for pf₂y, as follows:

       $p_{f2y} = p_{o1y} - p_{f1y} = 4.00kg*m*/s - 2.12 kg*m/s = 1.88 kg*m/s (10)$

• Since we have the x- and y- components of the final momentum of  the second particle, we can find its magnitude applying the Pythagorean Theorem, as follows:

       $p_{f2} = \sqrt{p_{f2x} ^{2} + p_{f2y} ^{2} } = \sqrt{(-0.12m/s)^{2} +(1.88m/s)^{2}} = 1.88 kg*m/s (11)$

• We can find the angle that this vector makes with the positive x- axis, applying the definition of tangent of an angle, as follows:

       $tg \theta = \frac{p_{2fy} }{p_{2fx} } = \frac{1.88m/s}{(-0.12m/s} = -15.7 (12)$

• The angle that we are looking for is just the arc tg of (12) which measured in a counter-clockwise direction from the positive x- axis, is just 273.4º.

b)

• Assuming that both masses are equal each other, we find that the momenta are proportional to the speeds, so we find that the relationship from the final kinetic energy and the initial one can be expressed as follows:

       $\frac{K_{f}}{K_{o} } = \frac{v_{f1}^{2} + v_{f2} ^{2}}{v_{o1}^{2} + v_{o2} ^{2} } = \frac{12.5}{20} = 0.63 (13)$

• So, the final kinetic energy has lost a 37% of the initial one.