Respuesta :
Answer:
1. The final velocity of the 30.0 g marble is 0.100 m/s to the left.
2. The final velocity of the 10.0 g marble is 0.500 m/s to the right.
3. The change in momentum for the 30.0 g marble is -9.00 × 10⁻³ kg · m/s
4. The change in momentum for the 10.0 g marble is 9.00 × 10⁻³ kg · m/s
5. The change in kinetic energy for the 30.0 g marble is -4.5 × 10⁻⁴ J
6. The change in kinetic energy for the 10.0 g marble is 4.5 × 10⁻⁴ J
Explanation:
Hi there!
Since the collision is elastic both the momentum and kinetic energy of the system comprised by the two marbles is conserved, i.e., it remains constant after the collision.
momentum before the collision = momentum after the collision
mA · vA + mB · vB = mA · vA´ + mB · vB´
Where:
mA and vA = mass and velocity of the 10.0 g marble.
mB and vB = mass and velocity of the 30.0 g marble.
vA´ and vB´ = final velocities of marble A and B respectively.
The kinetic energy of the system is also conserved:
kinetic energy before the collision = kinetic energy after the collision
1/2 mA · vA² + 1/2 mB · vB² = 1/2 mA · (vA´)² + 1/2 mB · (vB´)²
Then, replacing with the available data:
mA · vA + mB · vB = mA · vA´ + mB · vB´
0.010 kg · (-0.400 m/s) + 0.030 kg · 0.200 m/s = 0.010 kg · vA´ + 0.030 kg · vB´
2 × 10⁻³ kg · m/s = 0.010 kg · vA´ + 0.030 kg · vB´
Solving for vA´
0.2 kg · m/s - 3 kg · vB´ = vA´
Now, using conservation of the kinetic energy:
1/2 mA · vA² + 1/2 mB · vB² = 1/2 mA · (vA´)² + 1/2 mB · (vB´)²
0.010 kg · (-0.400 m/s)² + 0.030 kg · (0.200 m/s)² = 0.010 kg · (vA´)² + 0.030 kg · (vB´)²
2.8 × 10⁻³ kg · m/s = 0.010 kg · (vA´)² + 0.030 kg · (vB´)²
Replacing vA´:
2.8 × 10⁻³ kg · m/s = 0.010 kg · (vA´)² + 0.030 kg · (vB´)²
2.8 × 10⁻³ kg · m/s = 0.010 kg · (0.2 kg · m/s - 3 kg · vB´)² + 0.030 kg · (vB´)²
(I will omit units from this point for more clarity in the calculations)
2.8 × 10⁻³ = 0.010 (0.2 - 3 · vB´)² + 0.03 · (vB´)²
2.8 × 10⁻³ = 0.010(0.04 - 1.2 vB´ + 9(vB´)²) + 0.03(vB´)²
divide by 0.01 both sides of the equation:
0.28 = 0.04 - 1.2 vB´ + 9(vB´)² + 3(vB´)²
0 = -0.28 + 0.04 - 1.2 vB´ + 12(vB)²
0 = -0.24 - 1.2 vB´ + 12(vB)²
Solving the quadratic equation:
vB´= 0.200 m/s
vB´ = -0.100 m/s
The first value is discarded because it is the initial velocity. Then, the final velocity of the 30.0 g marble is 0.100 m/s to the left.
The velocity of the 10.0 g marble will be:
0.2 kg · m/s - 3 kg · vB´ = vA´
0.2 kg · m/s - 3 kg · (-0.100 m/s) = vA´
vA´ = 0.500 m/s
The final velocity of the 10.0 g marble is 0.500 m/s to the right.
The change in momentum of the 30.0 g marble is calculated as follows:
Δp = final momentum - initial momentum
Δp = 0.030 kg · (-0.100 m/s) -(0.030 kg · 0.200 m/s) = -9.00 × 10⁻³ kg · m/s
The change in momentum for the 30.0 g marble is -9.00 × 10⁻³ kg · m/s
The change in momentum of the 10.0 g marble is calculated in the same way:
Δp = final momentum - initial momentum
Δp = 0.010 kg · 0.500 m/s -(-0.010 kg · 0.400 m/s) = 9.00 × 10⁻³ kg · m/s
The change in momentum for the 10.0 g marble is 9.00 × 10⁻³ kg · m/s
The change in kinetic energy for the 30.0 g marble will be:
ΔKE = final kinetic energy - initial kinetic energy
ΔKE = 1/2 · 0.030 kg · (-0.100 m/s)² - 1/2 · 0.030 kg · (0.200 m/s)²
ΔKE = -4.5 × 10⁻⁴ J
The change in kinetic energy for the 30.0 g marble is -4.5 × 10⁻⁴ J
The change in kinetic energy for the 10.0 g marble will be:
ΔKE = final kinetic energy - initial kinetic energy
ΔKE = 1/2 · 0.010 kg · (0.500 m/s)² - 1/2 · 0.010 kg · (-0.400 m/s)²
ΔKE = 4.5 × 10⁻⁴ J
The change in kinetic energy for the 30.0 g marble is 4.5 × 10⁻⁴ J
The final velocity of the 10g marble after collision is 0.596 m/s to the right.
The final velocity of the 30g marble after collision is 0.00 4m/s to left.
The change in the momentum of the first marble (10g) is 0.01 kg.m/s.
The change in the momentum of the first marble (30g) is -0.01 kg.m/s.
The change in the kinetic energy of the 10 g marble is 0.00098 J.
The change in the kinetic energy of the 30 g marble is -0.0006 J.
The given parameters;
- mass of the first marble, m₁ = 10 g = 0.01 kg
- initial velocity of the first marble, u₁ = 0.4 m/s
- mass of the second marble, m₂ = 30 g
- initial velocity second, u₂ = 0.2 m/s
Apply the principle of conservation of linear momentum;
[tex]m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2\\\\(0.01)(-0.4) + (0.03 \times 0.2) = 0.01v_1 + 0.03v_2\\\\-0.004 + (0.006) = 0.01v_1+ 0.03v_2\\\\0.002 = 0.01v_1 + 0.03v_2[/tex]
where;
- v₁ is the final velocity of the first marble
- v₂ is the final velocity of the second marble
Apply one-directional velocity;
[tex]u_1 + v_1 = u_2 + v_2\\\\-0.4 + v_1 = 0.2+ v_2\\\\v_1 =0.6 + v_2[/tex]
Substitute the value of v₁ in the above equation;
[tex]0.002 = 0.01(0.6 + v_2) + v_2\\\\0.002 = 0.006 + 0.01v_2 + v_2 \\\\0.002 = 0.006 + 1.01v_2\\\\1.01v_2 = -0.004\\\\v_2 = \frac{-0.004}{1.01} \\\\v_2 = -0.004 \ m/s[/tex]
The final velocity of the first marble is calculated as follows;
[tex]v_1 = 0.6 - 0.004\\\\v_1 = 0.596 \ m/s[/tex]
The change in the momentum of the first marble (10g);
[tex]\Delta P = m_v_1 - m_1u_1\\\\\\Delta P = m_1(v_1 - u_1)\\\\\Delta P = 0.01(0.596 - -0.4)= 0.01(0.596 + 0.4)\\\\\Delta P =0.01 \ kg.m/s[/tex]
The change in the momentum of the second marble (30g);
[tex]\Delta P = m_2(v_2 - u_2)\\\\\Delta P = 0.03(-0.004 -0.2)\\\\\Delta P = -0.01 \ kg.m/s[/tex]
The change in the kinetic energy of the 10 g marble;
[tex]\Delta K.E_{10g} = \frac{1}{2} m_1(v_1^2 - u_1^2)\\\\\Delta K.E_{10g} = 0.5 \times 0.01(0.596^2 - 0.4^2)\\\\\Delta K.E_{10g} = 0.00098 \ J[/tex]
The change in the kinetic energy of the 30 g marble;
[tex]\Delta K.E_{30g} = \frac{1}{2} m_1(v_1^2 - u_1^2)\\\\\Delta K.E_{30g} = 0.5 \times 0.03(0.004^2 - 0.2^2)\\\\\Delta K.E_{30g} = - 0.0006 \ J[/tex]
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