Question

two balls x and y with the same diameter are fired horizontally with the same initial momentum from the same height above the ground. the mass of x is greater than the mass of y. air resistance is negligible. what is correct about the horizontal distances travelled by x and y and the times taken by x and y to reach the ground? horizontal distances time to reach ground a x and y the same x and y times the same b x and y the same x takes a shorter time than y c x less than y x and y times the same d x less than y x takes a shorter time than y

Explanation:

Step1: Analyze time - of - flight

The vertical motion of the two balls is a free - fall motion. The initial vertical velocity \(u_y = 0\) for both balls, and the vertical displacement \(y=-h\) (taking downwards as negative), and the acceleration \(a=-g\). Using the equation \(y = u_y t+\frac{1}{2}at^{2}\), substituting \(u_y = 0\) and \(a=-g\) and \(y = - h\), we get \(-h=0\times t-\frac{1}{2}gt^{2}\), or \(t=\sqrt{\frac{2h}{g}}\). Since \(h\) and \(g\) are the same for both balls, the time taken for \(X\) and \(Y\) to reach the ground is the same.

Step2: Analyze horizontal distance

The initial momentum \(p = mu\) is the same for both balls, and \(m_X>m_Y\). So, from \(p = mu\), the initial horizontal velocity \(u=\frac{p}{m}\), which means \(u_X m_Y\)). The horizontal distance \(x = u_x t\). Since \(t\) is the same for both and \(u_X

Answer:

C. X less than Y, X and Y times the same