I don't know that; sorry, I should just be removed from here.
Answer:
All three pendulums will have the same angular frequencies.
Explanation:
For a simple pendulum, the time period using the approximation 
is expressed as:
The angular frequency 
is defined as
Since the angular frequency remains unaffected by the initial angle (valid strictly for small angle approximations), we deduce that the angular frequencies of the three pendulums are identical.
<span>We will apply the momentum-impulse theorem here. The total momentum along the x-direction is defined as p_(f) = p_(1) + p_(2) + p_(3) = 0.
Therefore, p_(1x) = m1v1 = 0.2 * 2 = 0.4. Additionally, p_(2x) = m2v2 = 0 and p_(3x) = m3v3 = 0.1 *v3, where v3 represents the unknown speed and m3 signifies the mass of the third object, which has an unspecified velocity.
In the same way, for the particle of 235g, the y-component of the total momentum is described with p_(fy) = p_(1y) + p_(2y) + p_(3y) = 0.
Thus, p_(1y) = 0, p_(2y) = m2v2 = 0.235 * 1.5 = 0.3525 and p_(3y) = m3v3 = 0.1 * v3, where m3 is the mass of the third piece.
Consequently, p_(fx) = p_(1x) + p_(2x) + p_(3x) = 0.4 + 0.1v3; yielding v3 = 0.4/-0.1 = - 4.
Similarly, p_(fy) = 0.3525 + 0.1v3; thus v3 = - 0.3525/0.1 = -3.525.
Therefore, the x-component of the speed of the third piece is v_3x = -4 and the y-component is v_3y = 3.525.
The overall speed is calculated as follows: resultant = âš (-4)^2 + (-3.525)^2 = 5.335</span>
