To determine the velocity of the truck after the collision, we can use the principle of conservation of momentum, considering it is an elastic collision.
The conservation of momentum states:
\[ m_1 v_{1i} + m_2 v_{2i} = m_1 v_{1f} + m_2 v_{2f} \]
Where:
- \( m_1 \) = mass of the car = \( 1.1 \times 10^3 \) kg
- \( v_{1i} \) = initial velocity of the car = \( +22.0 \) m/s
- \( v_{1f} \) = final velocity of the car = \( -11.0 \) m/s
- \( m_2 \) = mass of the truck = \( 2.3 \times 10^3 \) kg
- \( v_{2i} \) = initial velocity of the truck = \( 0 \) m/s (stationary)
- \( v_{2f} \) = final velocity of the truck (unknown)
Plugging the values into the equation:
\[ (1.1 \times 10^3 , \text{kg})(22.0 , \text{m/s}) + (2.3 \times 10^3 , \text{kg})(0 , \text{m/s}) = (1.1 \times 10^3 , \text{kg})(-11.0 , \text{m/s}) + (2.3 \times 10^3 , \text{kg})(v_{2f}) \]
Calculating the left side:
\[ (1.1 \times 10^3)(22.0) = 24200 , \text{kg m/s} \]
Calculating the right side:
\[ (1.1 \times 10^3)(-11.0) = -12100 , \text{kg m/s} \]
Now substituting these values back into the equation:
\[ 24200 = -12100 + (2.3 \times 10^3)(v_{2f}) \]
Adding \( 12100 \) to both sides:
\[ 24200 + 12100 = (2.3 \times 10^3)(v_{2f}) \]
This simplifies to:
\[ 36300 = (2.3 \times 10^3)(v_{2f}) \]
Now, solving for \( v_{2f} \):
\[ v_{2f} = \frac{36300}{2.3 \times 10^3} \]
Calculating this gives:
\[ v_{2f} = \frac{36300}{2300} \approx 15.8 , \text{m/s} \]
The closest answer option is:
C. +15.7 meters/second.
1 answer