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3.10.12 deux rondelles rebondissent l’une sur l’autre, en deux dimensions. deux rondelles entrent en collision sur une surface horizon- tale. immédiatement avant la collision, la rondelle a, dont la masse est de 500 g, se déplace à 3 m/s vers l’est et la rondelle b, dont la masse de 800 g, se déplace à 4 m/s vers le sud (schéma ci-contre). immédiatement après la collision, la rondelle a se déplace à 2 m/s vers le sud. déterminez (a) le module et l’orientation de la vitesse de la rondelle b immédiatement après la collision, et (b) le pourcentage d’énergie cinétique perdue lors de la collision. 3.10.13 deux rondelles restent collées ensemble, en deux dimensions. dans la situation de l’exercice 3.10.12, on place un...
To solve this two - dimensional collision problem, we will use the principle of conservation of momentum.
Part (a): Determine the magnitude and direction of the velocity of puck B immediately after the collision
Let's establish a coordinate system. Let the north - south direction be the y - axis (with north as the positive y - direction) and the east - west direction be the x - axis (with east as the positive x - direction).
Step 1: Analyze the momentum before the collision
- Puck A: Mass \(m_A = 500\space g=0.5\space kg\), initial velocity \(v_{Aix} = 0\) (since it is moving along the y - axis), \(v_{Aiy}=- 3\space m/s\) (negative because it is moving south)
- Puck B: Mass \(m_B = 800\space g = 0.8\space kg\), initial velocity \(v_{Bix}=-4\space m/s\) (negative because it is moving west), \(v_{Biy} = 0\) (since it is moving along the x - axis)
The initial momentum in the x - direction \(P_{ix}=m_Av_{Aix}+m_Bv_{Bix}\)
\(P_{ix}=0.5\times0 + 0.8\times(- 4)=-3.2\space kg\cdot m/s\)
The initial momentum in the y - direction \(P_{iy}=m_Av_{Aiy}+m_Bv_{Biy}\)
\(P_{iy}=0.5\times(-3)+0.8\times0=- 1.5\space kg\cdot m/s\)
Step 2: Analyze the momentum after the collision
After the collision, the two pucks stick together. Let the mass of the combined system be \(M=m_A + m_B=0.5 + 0.8 = 1.3\space kg\)
Let the velocity of the combined system be \(\vec{v}=v_x\hat{i}+v_y\hat{j}\)
The final momentum in the x - direction \(P_{fx}=Mv_x\)
The final momentum in the y - direction \(P_{fy}=Mv_y\)
By the law of conservation of momentum, \(P_{ix}=P_{fx}\) and \(P_{iy}=P_{fy}\)
For the x - direction:
\(Mv_x=P_{ix}\)
\(v_x=\frac{P_{ix}}{M}=\frac{-3.2}{1.3}\approx - 2.46\space m/s\)
For the y - direction:
\(Mv_y=P_{iy}\)
\(v_y=\frac{P_{iy}}{M}=\frac{-1.5}{1.3}\approx - 1.15\space m/s\)
Now, we need to find the velocity of puck B before the collision and the velocity of the combined system after the collision to find the velocity of puck B after the collision? Wait, no, the problem says "Deux rondelles rebondissent l'une sur l'autre, en deux dimensions. Deux rondelles entrent en collision sur une surface horizontale. Immédiatement avant la collision, la rondelle A, dont la masse est de 500 g, se déplace à 3 m/s vers l'ouest et la rondelle B, dont la masse de 800 g, se déplace à 4 m/s vers le sud (schéma ci - contre). Immédiatement après la collision, la rondelle A se déplace à 2 m/s vers le sud. Déterminez (a) le module et l'orientation de la vitesse de la rondelle B immédiatement après la collision, et (b) le pourcentage d'énergie cinétique perdue lors de la collision."
Let's re - do the momentum conservation correctly. Let's define:
- For puck A: \(m_A = 0.5\space kg\), \(\vec{v}_{A_i}=- 3\hat{j}\space m/s\) (west is negative y? Wait, maybe I mixed up the axes. Let's re - define: Let the positive x - axis be east and positive y - axis be north. So, initial velocity of A: \(v_{Ax_i}=0\), \(v_{Ay_i}=- 3\space m/s\) (south is negative y)
Initial velocity of B: \(v_{Bx_i}=-4\space m/s\) (west is negative x), \(v_{By_i}=0\)
After collision:
Velocity of A: \(v_{Ax_f}=0\), \(v_{Ay_f}=-2\space m/s\) (south is negative y)
Let the velocity of B be \(\vec{v}_B = v_{Bx}\hat{i}+v_{By}\hat{j}\)
By conservation of momentum in x - direction:
\(m_Av_{Ax_i}+m_Bv_{Bx_i}=m_Av_{Ax_f}+m_Bv_{Bx}\)
\(0.5\times0 + 0.8\times(-4)=0.5\times0+0.8\times v_{Bx}\)
\(-3.2 = 0.8v_{Bx}\)
\(v_{Bx}=\frac{-3.2}{0.8}=-4\space m/s\)
By conservation of momentum in y - direction:
\(m_Av_{Ay_i}+m_Bv_{By_i}=m_Av_{Ay_f}+m_Bv_{By}\)
\(0.5\times(-3)+0.8\times0 = 0.5\times(-2)+0.8\…
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To solve this two - dimensional collision problem, we will use the principle of conservation of momentum.
Part (a): Determine the magnitude and direction of the velocity of puck B immediately after the collision
Let's establish a coordinate system. Let the north - south direction be the y - axis (with north as the positive y - direction) and the east - west direction be the x - axis (with east as the positive x - direction).
Step 1: Analyze the momentum before the collision
- Puck A: Mass \(m_A = 500\space g=0.5\space kg\), initial velocity \(v_{Aix} = 0\) (since it is moving along the y - axis), \(v_{Aiy}=- 3\space m/s\) (negative because it is moving south)
- Puck B: Mass \(m_B = 800\space g = 0.8\space kg\), initial velocity \(v_{Bix}=-4\space m/s\) (negative because it is moving west), \(v_{Biy} = 0\) (since it is moving along the x - axis)
The initial momentum in the x - direction \(P_{ix}=m_Av_{Aix}+m_Bv_{Bix}\)
\(P_{ix}=0.5\times0 + 0.8\times(- 4)=-3.2\space kg\cdot m/s\)
The initial momentum in the y - direction \(P_{iy}=m_Av_{Aiy}+m_Bv_{Biy}\)
\(P_{iy}=0.5\times(-3)+0.8\times0=- 1.5\space kg\cdot m/s\)
Step 2: Analyze the momentum after the collision
After the collision, the two pucks stick together. Let the mass of the combined system be \(M=m_A + m_B=0.5 + 0.8 = 1.3\space kg\)
Let the velocity of the combined system be \(\vec{v}=v_x\hat{i}+v_y\hat{j}\)
The final momentum in the x - direction \(P_{fx}=Mv_x\)
The final momentum in the y - direction \(P_{fy}=Mv_y\)
By the law of conservation of momentum, \(P_{ix}=P_{fx}\) and \(P_{iy}=P_{fy}\)
For the x - direction:
\(Mv_x=P_{ix}\)
\(v_x=\frac{P_{ix}}{M}=\frac{-3.2}{1.3}\approx - 2.46\space m/s\)
For the y - direction:
\(Mv_y=P_{iy}\)
\(v_y=\frac{P_{iy}}{M}=\frac{-1.5}{1.3}\approx - 1.15\space m/s\)
Now, we need to find the velocity of puck B before the collision and the velocity of the combined system after the collision to find the velocity of puck B after the collision? Wait, no, the problem says "Deux rondelles rebondissent l'une sur l'autre, en deux dimensions. Deux rondelles entrent en collision sur une surface horizontale. Immédiatement avant la collision, la rondelle A, dont la masse est de 500 g, se déplace à 3 m/s vers l'ouest et la rondelle B, dont la masse de 800 g, se déplace à 4 m/s vers le sud (schéma ci - contre). Immédiatement après la collision, la rondelle A se déplace à 2 m/s vers le sud. Déterminez (a) le module et l'orientation de la vitesse de la rondelle B immédiatement après la collision, et (b) le pourcentage d'énergie cinétique perdue lors de la collision."
Let's re - do the momentum conservation correctly. Let's define:
- For puck A: \(m_A = 0.5\space kg\), \(\vec{v}_{A_i}=- 3\hat{j}\space m/s\) (west is negative y? Wait, maybe I mixed up the axes. Let's re - define: Let the positive x - axis be east and positive y - axis be north. So, initial velocity of A: \(v_{Ax_i}=0\), \(v_{Ay_i}=- 3\space m/s\) (south is negative y)
Initial velocity of B: \(v_{Bx_i}=-4\space m/s\) (west is negative x), \(v_{By_i}=0\)
After collision:
Velocity of A: \(v_{Ax_f}=0\), \(v_{Ay_f}=-2\space m/s\) (south is negative y)
Let the velocity of B be \(\vec{v}_B = v_{Bx}\hat{i}+v_{By}\hat{j}\)
By conservation of momentum in x - direction:
\(m_Av_{Ax_i}+m_Bv_{Bx_i}=m_Av_{Ax_f}+m_Bv_{Bx}\)
\(0.5\times0 + 0.8\times(-4)=0.5\times0+0.8\times v_{Bx}\)
\(-3.2 = 0.8v_{Bx}\)
\(v_{Bx}=\frac{-3.2}{0.8}=-4\space m/s\)
By conservation of momentum in y - direction:
\(m_Av_{Ay_i}+m_Bv_{By_i}=m_Av_{Ay_f}+m_Bv_{By}\)
\(0.5\times(-3)+0.8\times0 = 0.5\times(-2)+0.8\times v_{By}\)
\(-1.5=-1 + 0.8v_{By}\)
\(0.8v_{By}=-1.5 + 1=-0.5\)
\(v_{By}=\frac{-0.5}{0.8}=-0.625\space m/s\)
The magnitude of the velocity of B after collision is \(v_B=\sqrt{v_{Bx}^2 + v_{By}^2}=\sqrt{(-4)^2+(-0.625)^2}=\sqrt{16 + 0.390625}=\sqrt{16.390625}\approx4.05\space m/s\)
The direction \(\theta\) (measured from the negative x - axis towards the negative y - axis) is given by \(\tan\theta=\frac{\vert v_{By}\vert}{\vert v_{Bx}\vert}=\frac{0.625}{4}=0.15625\)
\(\theta=\arctan(0.15625)\approx9.0^{\circ}\) below the west direction (or \(180^{\circ}+9.0^{\circ}=189.0^{\circ}\) from the positive x - axis)
Part (b): Determine the percentage of kinetic energy lost
Step 1: Calculate the initial kinetic energy \(K_i\)
The initial kinetic energy is the sum of the kinetic energies of A and B before the collision.
\(K_i=\frac{1}{2}m_Av_{A_i}^2+\frac{1}{2}m_Bv_{B_i}^2\)
\(v_{A_i}=3\space m/s\), \(v_{B_i}=4\space m/s\)
\(K_i=\frac{1}{2}\times0.5\times3^2+\frac{1}{2}\times0.8\times4^2\)
\(K_i = 0.25\times9+0.4\times16\)
\(K_i=2.25 + 6.4=8.65\space J\)
Step 2: Calculate the final kinetic energy \(K_f\)
The final kinetic energy is the sum of the kinetic energies of A and B after the collision.
\(K_f=\frac{1}{2}m_Av_{A_f}^2+\frac{1}{2}m_Bv_{B}^2\)
\(v_{A_f}=2\space m/s\), \(v_B\approx4.05\space m/s\)
\(K_f=\frac{1}{2}\times0.5\times2^2+\frac{1}{2}\times0.8\times(4.05)^2\)
\(K_f=0.25\times4+0.4\times16.4025\)
\(K_f = 1+6.561=7.561\space J\)
Step 3: Calculate the percentage of kinetic energy lost
The kinetic energy lost \(\Delta K=K_i - K_f\)
\(\Delta K=8.65 - 7.561 = 1.089\space J\)
The percentage of kinetic energy lost is \(\frac{\Delta K}{K_i}\times100=\frac{1.089}{8.65}\times100\approx12.6\%\)
Part (a) Answer
The magnitude of the velocity of puck B after the collision is approximately \(4.05\space m/s\) and the direction is approximately \(9.0^{\circ}\) south of west (or \(189.0^{\circ}\) from the positive x - axis)
Part (b) Answer
The percentage of kinetic energy lost is approximately \(12.6\%\)