a ball moving witha speed of 9m/s sttrikes an identical ball such that after collison ,the direction of each ball makes an angle of 30 with original line of motion. if K1 and K2 are the kinetic energies before and after collision the A) K1< K2 b) K1> K2 c)K1=K2
To determine if the kinetic energy after the collision (K2) is greater, less or equal to the kinetic energy before the collision (K1), we need to apply the principles of conservation of momentum and conservation of kinetic energy.
1. First, let's consider the conservation of momentum. In an elastic collision, the total momentum before and after the collision should be the same. The momentum of an object is defined as its mass multiplied by its velocity. Since the masses of the two identical balls are the same, we only need to consider the velocities.
Let v1 be the velocity of the first ball after the collision and v2 be the velocity of the second ball after the collision. Since the directions of both balls make an angle of 30 degrees with the original line of motion, they form an isosceles triangle as shown:
/
v2 / _(30°)_
/ \
------------
\ /
\ / (30°) v1
2. From the given information, we know the initial velocity of both balls is 9 m/s.
3. Using trigonometry, we can determine the components of the velocities parallel to the original line of motion:
v1_parallel = v1 * cos(30°)
v2_parallel = v2 * cos(30°)
4. The law of conservation of momentum states that the total momentum before and after the collision should be equal:
m * v1_initial + m * v2_initial = m * v1_parallel + m * v2_parallel
Since the masses cancel out, we're left with:
v1_initial + v2_initial = v1_parallel + v2_parallel
Assuming the initial velocities of both balls are the same (9 m/s), we can simplify the equation:
2 * 9m/s = v1_parallel + v2_parallel
18m/s = v1_parallel + v2_parallel
5. Now, let's consider the conservation of kinetic energy. The kinetic energy is given by the equation:
KE = (1/2) * m * v^2
The kinetic energy before the collision is:
K1 = (1/2) * m * v1_initial^2
And the kinetic energy after the collision is:
K2 = (1/2) * m * v1_parallel^2 + (1/2) * m * v2_parallel^2
Plugging in the values, we get:
K2 = (1/2) * v1_parallel^2 + (1/2) * v2_parallel^2
6. To compare K1 and K2, we need to simplify further:
K2 = (1/2) * (v1_parallel^2 + v2_parallel^2)
= (1/2) * [(v1_initial * cos(30°))^2 + (v2_initial * cos(30°))^2]
= (1/2) * [(9m/s * cos(30°))^2 + (9m/s * cos(30°))^2]
= (1/2) * [9m/s * cos(30°)]^2 * (cos^2(30°) + cos^2(30°))
= (1/2) * [9m/s * cos(30°)]^2 * 2 * cos^2(30°)
= 9m/s * cos^2(30°) * [9m/s * cos(30°)]
Note: We can use the identity cos^2(θ) = (1 + cos(2θ)) / 2.
K2 = 9m/s * cos^2(30°) * [9m/s * cos(30°)]
= 9m/s * (1/2) * [9m/s + 9m/s * cos(60°)]
7. Simplifying further:
K2 = 9m/s * (1/2) * [9m/s + 9m/s * (1/2)]
= 9m/s * (1/2) * [9m/s + 4.5m/s]
= 9m/s * (1/2) * 13.5m/s
= 58.5m^2/s^2
8. Now, let's compare K1 and K2:
K1 = (1/2) * m * v1_initial^2
= (1/2) * 9m/s * 9m/s
= 40.5m^2/s^2
By comparing K1 and K2, we find that K1 < K2. Therefore, the correct answer is A) K1 < K2.