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Bumper car \( 1(281 \mathrm{~kg}) \) moves east at \( 2.82 \mathrm{~m} / \mathrm{s} \). Bumper car \( 2(209 \mathrm{~kg}) \) moves west at \( 1.72 \mathrm{~m} / \mathrm{s} \). If the cars have an elastic collision, what would the velocity of car 1 be afterwards? \[ \mathrm{v}_{1 \mathrm{f}}=[?] \mathrm{m} / \mathrm{s} \] Remember: East is positive \( (+) \); west is negative \( (-) \).

Ask by Curry Chandler. in the United States
Mar 19,2025

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Answer

After the elastic collision, bumper car 1 moves west at approximately \( -1.05 \, \mathrm{m/s} \).

Solution

The formula for the final velocity of car 1 in an elastic head‑on collision is \[ v_{1f}=\frac{m_1-m_2}{m_1+m_2}\,v_{1i}+\frac{2m_2}{m_1+m_2}\,v_{2i}. \] Given: - \( m_1=281\,\mathrm{kg} \) - \( v_{1i}=2.82\,\mathrm{m/s} \) (east, so positive) - \( m_2=209\,\mathrm{kg} \) - \( v_{2i}=-1.72\,\mathrm{m/s} \) (west, so negative) Step 1: Compute the sum and difference of the masses \[ m_1+m_2 = 281+209 = 490\,\mathrm{kg}, \] \[ m_1-m_2 = 281-209 = 72\,\mathrm{kg}. \] Step 2: Substitute the values into the formula \[ v_{1f}=\frac{72}{490}\,(2.82)+\frac{2(209)}{490}\,(-1.72). \] Step 3: Simplify the fractions \[ \frac{72}{490}\approx0.14694, \] \[ 2(209)=418,\quad \frac{418}{490}\approx0.85204. \] Step 4: Calculate each term First term: \[ 0.14694\times2.82\approx0.4146\,\mathrm{m/s}, \] Second term: \[ 0.85204\times(-1.72)\approx-1.4655\,\mathrm{m/s}. \] Step 5: Add the terms \[ v_{1f}=0.4146+(-1.4655)\approx-1.0509\,\mathrm{m/s}. \] Thus, the final velocity of car 1 is approximately \[ v_{1f}\approx-1.05\,\mathrm{m/s}. \] Since east is defined as positive, the negative sign indicates that car 1 moves west after the collision.

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To find the final velocity of bumper car \( 1 \) after the elastic collision, we can use the principles of conservation of momentum and conservation of kinetic energy. 1. First, we assign the velocities: - Initial velocity of car \( 1 \), \( v_{1i} = +2.82 \, \text{m/s} \) (east). - Initial velocity of car \( 2 \), \( v_{2i} = -1.72 \, \text{m/s} \) (west). 2. The total momentum before the collision can be calculated as: \[ p_{initial} = m_1 v_{1i} + m_2 v_{2i} = (281 \, \text{kg})(2.82 \, \text{m/s}) + (209 \, \text{kg})(-1.72 \, \text{m/s}) \] 3. The total kinetic energy before the collision is: \[ KE_{initial} = \frac{1}{2} m_1 v_{1i}^2 + \frac{1}{2} m_2 v_{2i}^2 \] 4. After the collision, we assume \( v_{1f} \) and \( v_{2f} \) are the final velocities of cars \( 1 \) and \( 2 \). 5. We can set up the equations based on the conservation laws: \[ m_{1} v_{1i} + m_{2} v_{2i} = m_{1} v_{1f} + m_{2} v_{2f} \] \[ \frac{1}{2} m_{1} v_{1i}^2 + \frac{1}{2} m_{2} v_{2i}^2 = \frac{1}{2} m_{1} v_{1f}^2 + \frac{1}{2} m_{2} v_{2f}^2 \] 6. Solving these equations gives us the final velocities. However, the final velocities after an elastic collision can also be calculated using a derived formula for elastic collisions: \[ v_{1f} = \frac{(m_{1} - m_{2}) v_{1i} + 2 m_{2} v_{2i}}{m_{1} + m_{2}} \] Substituting the values into this formula: \[ v_{1f} = \frac{(281 - 209)(2.82) + 2 \cdot 209(-1.72)}{281 + 209} \] \[ v_{1f} = \frac{72(2.82) - 718.96}{490} \] Calculating this gives \( v_{1f} \). After performing these calculations, we find that: \[ v_{1f} \approx 0.06 \, \text{m/s} \] Thus, the velocity of car \( 1 \) after the collision is: \[ \mathrm{v}_{1 \mathrm{f}} \approx 0.06 \, \mathrm{m/s} \]

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