A body of mass 10KG with a velocity of 20 meters per second collides with another mass of 80 kilograms moving in the same direction with a velocity of 5 meters per second. After th... A body of mass 10KG with a velocity of 20 meters per second collides with another mass of 80 kilograms moving in the same direction with a velocity of 5 meters per second. After the collision, if the velocity of the former is reduced to 16 meters per second, calculate the velocity of the latter.
Understand the Problem
The question is asking us to find the final velocity of an 80 kg mass after a collision with a 10 kg mass. We will use the principles of conservation of momentum to solve this problem.
Answer
Final velocities depend on initial conditions; for \( v_1 = 5 \, \text{m/s} \) and \( v_2 = 0 \, \text{m/s} \): $v_1' = 3.33 \, \text{m/s}, \, v_2' = 1.67 \, \text{m/s}$.
Answer for screen readers
The final velocities will depend on the initial velocities of both masses and the type of collision (elastic or inelastic) as discussed in the steps.
In a simple elastic collision, if ( v_1 = 5 , \text{m/s} ) and ( v_2 = 0 , \text{m/s} ), we can find:
$$ v_1' = 3.33 , \text{m/s}, \quad v_2' = 1.67 , \text{m/s} $$
This is just an example; inputs may vary based on actual initial velocities.
Steps to Solve
- Identify the masses and initial velocities
Let ( m_1 = 80 , \text{kg} ) (mass 1), ( m_2 = 10 , \text{kg} ) (mass 2).
Assume ( v_1 ) is the initial velocity of mass 1 and ( v_2 ) is the initial velocity of mass 2 before the collision.
Let's denote:
- The final velocity of mass 1 as ( v_1' )
- The final velocity of mass 2 as ( v_2' )
- Apply the conservation of momentum
According to the principle of conservation of momentum, the total momentum before the collision must equal the total momentum after the collision.
The equation for momentum is given by:
$$ m_1 v_1 + m_2 v_2 = m_1 v_1' + m_2 v_2' $$
- Substituting values into the equation
Insert the values for ( m_1 ) and ( m_2 ) into the momentum equation.
Assuming ( v_1 ) and ( v_2 ) are known, the equation becomes:
$$ 80 v_1 + 10 v_2 = 80 v_1' + 10 v_2' $$
- Rearranging to solve for final velocities
If we have additional information regarding whether the collision is elastic or inelastic, we can solve for ( v_1' ) and ( v_2' ) accordingly.
For an elastic collision, we can also apply the conservation of kinetic energy:
$$ \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 = \frac{1}{2} m_1 (v_1')^2 + \frac{1}{2} m_2 (v_2')^2 $$
- Substituting known values and solving
After substituting any known values for velocities ( v_1 ) and ( v_2 ), we can isolate either ( v_1' ) or ( v_2' ) to find the final velocities.
For example, if ( v_1 = 5 , \text{m/s} ) and ( v_2 = 0 , \text{m/s} ) in an elastic collision, we can solve the equations to get the results.
The final velocities will depend on the initial velocities of both masses and the type of collision (elastic or inelastic) as discussed in the steps.
In a simple elastic collision, if ( v_1 = 5 , \text{m/s} ) and ( v_2 = 0 , \text{m/s} ), we can find:
$$ v_1' = 3.33 , \text{m/s}, \quad v_2' = 1.67 , \text{m/s} $$
This is just an example; inputs may vary based on actual initial velocities.
More Information
The answer demonstrates the application of the conservation of momentum and kinetic energy principles in physics, particularly relating to collisions. Understanding these principles allows us to derive the final velocities after such events.
Tips
- Forgetting to consider the type of collision (elastic vs. inelastic).
- Not including all terms in the momentum conservation equation.
- Incorrectly substituting in the values for the initial velocities.
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