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2.A billiard ball at rest is struck by the cue ball of the same mass whose speed is 5 m/s .After an elastic collision the cue ball goes off at 50 degree with respect to its original direction of motion and the struck ball goes off at 40 degree with respect to this direction,What are the final speed of the balls?Illustrate the situation given
Final answer:
v1 = 3.83 m/s
v2 = 3.21 m/s
To solve this problem, we can use the principles of conservation of momentum and conservation of kinetic energy. Let's break down the problem step by step:
1. Draw a diagram to illustrate the situation given. Label the initial and final directions of motion for both balls.
2. Apply the conservation of momentum. Since momentum is conserved in elastic collisions, the total momentum before the collision must be equal to the total momentum after the collision.
In this case, we have two balls of the same mass, so their momenta can be expressed as:
m1 * v1 = m2 * v2
Where m1 and m2 are the masses of the balls and v1 and v2 are their respective velocities.
3. Apply the conservation of kinetic energy. In an elastic collision, the total kinetic energy before the collision must be equal to the total kinetic energy after the collision.
In this case, the initial kinetic energy is given by:
(1/2) * m1 * v1^2 + (1/2) * m2 * 0^2 (since the second ball is at rest)
The final kinetic energy is given by:
(1/2) * m1 * v1'^2 + (1/2) * m2 * v2'^2
Note that the prime symbol (') indicates the final velocities.
4. Use the given information to find the angles and velocities. You are given that the cue ball goes off at a 50-degree angle and the struck ball goes off at a 40-degree angle. Let's assume the original direction of motion is along the positive x-axis.
Based on these angles, we can find the final velocities using trigonometry:
v1' = v1 * cos(50°)
v2' = v1 * sin(50°) + 0 (since the second ball is initially at rest)
5. Plug the final velocities into the equations from step 3 and solve for the final speeds.
(1/2) * m1 * v1^2 + (1/2) * m2 * 0^2 = (1/2) * m1 * v1'^2 + (1/2) * m2 * v2'^2
Simplify the equation and solve for v1'^2 and v2'^2. Then, take the square root to find the final speeds of the balls, v1' and v2'.
Now you have the final speeds of the balls after the collision.
I what an anwer
v1fcos40+v2fcos50=5
v1fsin40 = v2fsin50
solve 2nd eq for v1f or vsf and plug into eq 1