A barge with mass 1.7×105 kg is proceeding down river at 6.1 m/s in heavy fog when it collides broadside with a barge heading directly across the river. The second barge has mass 2.83×105 kg and was moving at 4.2 m/s. Immediately after impact, the second barge finds its course deflected by 18° in the downriver direction and its speed increased to 5.0 m/s. The river current was practically zero at the time of the accident. What are the speed and direction of motion of the first barge immediately after the collision? a)Speed?
b)Enter the angle of deflection from the original direction with + as motion to its left and - as to its right.
c)How much kinetic energy is lost in the collision?
Use the law of conservation of momentum to solve this. Since that is a vector, do it in both directions (crossways).
I used the equation m1vf-m1vi=-m2vf+m2vi and got a velocity of 4.77m/s but that answer is incorrect.
Am I not using the right formula for question (a)?
ok i figured out what I did wrong for the (a). I also found (b). But I cannot seem to get the right answer for the kinetic energy lost in the collision.
how do you do part a?
To solve this problem, we need to apply the principles of conservation of momentum and conservation of kinetic energy.
a) To find the speed of the first barge immediately after the collision, we can use the equation:
(m1 * v1) + (m2 * v2) = (m1 * v1f) + (m2 * v2f)
where
m1 = mass of the first barge = 1.7 × 10^5 kg
v1 = initial velocity of the first barge = 6.1 m/s
m2 = mass of the second barge = 2.83 × 10^5 kg
v2 = initial velocity of the second barge = 4.2 m/s
v1f = final velocity of the first barge (unknown)
v2f = final velocity of the second barge = 5.0 m/s
Plugging in the values, the equation becomes:
(1.7 × 10^5 kg * 6.1 m/s) + (2.83 × 10^5 kg * 4.2 m/s) = (1.7 × 10^5 kg * v1f) + (2.83 × 10^5 kg * 5.0 m/s)
Simplifying the equation gives us:
1.03 × 10^6 kg·m/s + 1.19 × 10^6 kg·m/s = (1.7 × 10^5 kg * v1f) + (1.415 × 10^6 kg·m/s)
Combining like terms:
2.22 × 10^6 kg·m/s = (1.7 × 10^5 kg * v1f) + (1.415 × 10^6 kg·m/s)
Rearranging the equation and isolating v1f:
(1.7 × 10^5 kg * v1f) = 2.22 × 10^6 kg·m/s - (1.415 × 10^6 kg·m/s)
v1f = (2.22 × 10^6 kg·m/s - 1.415 × 10^6 kg·m/s) / (1.7 × 10^5 kg)
v1f = 8052.94 m/s
Therefore, the speed of the first barge immediately after the collision is approximately 8052.94 m/s.
b) To find the angle of deflection from the original direction, we can use the equation:
θ = tan^(-1)((v2f * sin(θ2))/(v1f + v2f * cos(θ2)))
where
θ = angle of deflection from the original direction (unknown)
v2f = final velocity of the second barge = 5.0 m/s
θ2 = angle of deflection of the second barge = 18°
Plugging in the values, the equation becomes:
θ = tan^(-1)((5.0 m/s * sin(18°))/(8052.94 m/s + 5.0 m/s * cos(18°)))
Evaluating this equation gives us:
θ ≈ -0.21°
Therefore, the angle of deflection from the original direction is approximately -0.21°, indicating a slight deflection to the right.
c) To find the amount of kinetic energy lost in the collision, we can calculate the initial kinetic energy (KE_initial) and final kinetic energy (KE_final), and then find the difference between them.
KE_initial = 0.5 * m1 * v1^2 + 0.5 * m2 * v2^2
KE_final = 0.5 * m1 * v1f^2 + 0.5 * m2 * v2f^2
Using the given values, we have:
KE_initial = 0.5 * 1.7 × 10^5 kg * (6.1 m/s)^2 + 0.5 * 2.83 × 10^5 kg * (4.2 m/s)^2
KE_final = 0.5 * 1.7 × 10^5 kg * (8052.94 m/s)^2 + 0.5 * 2.83 × 10^5 kg * (5.0 m/s)^2
Calculating these values gives us:
KE_initial ≈ 1.36 × 10^8 J
KE_final ≈ 1.38 × 10^8 J
The amount of kinetic energy lost in the collision can be found by subtracting the final kinetic energy from the initial kinetic energy:
KE_lost = KE_initial - KE_final
≈ 1.36 × 10^8 J - 1.38 × 10^8 J
≈ -2 × 10^6 J
Therefore, the kinetic energy lost in the collision is approximately 2 × 10^6 J.