A1 kg ball is hung at the end of a rod 1 m long. if the system balances at a point on the rod one third of the distance from the end holding the mass, what is the mass of the rod?

The mass of the rod on which balances the ball hung at one end is 2 kg.

The given parameters;

The one-third distance from the point holding the mass is calculated as follows;

p = ¹/₃ x 1 = 0.33 m

0                                 0.33 m                                                        1 m

------------------------------Δ----------------------------------------------------------

↓                                                                                                      ↓

W                                                                                                 1 kg

Take moment about the pivot;

W(0.33) = 1(1 - 0.33)

W(0.33) = 1(0.67)

Thus, the mass of the rod on which balances the ball hung at one end is 2 kg.

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2 kg

Explanation:

Assuming the rod's mass is uniformly distributed, the center of mass is at half the length.

Sum of the moments at the balance point:

-(Mg)(L/3) + (mg)(L/2 − L/3) = 0

(Mg)(L/3) = (mg)(L/2 − L/3)

(Mg)(L/3) = (mg)(L/6)

2M = m

M = 1 kg, so m = 2 kg.

The mass of the rod is 2 kg.

B. 50 m/s²

Explanation:

Given:

Initial velocity of the ball is,

Final velocity of the ball is,

Time taken by the  ball is,

Acceleration is defined as the rate of change of velocity. So, the velocity of the ball is changing with time. So, acceleration is given as:

Plug in 0 for 'u', 45 for 'v', 0.9 for 't' and solve for acceleration, 'a'. This gives,

Therefore, the acceleration of the ball is 50 m/s².