Into a small metal plate weighing m = 0.2 kg. suspended on a thread with a length of l = 1 m, a ball of mass m

Into a small metal plate weighing m = 0.2 kg. suspended on a thread with a length of l = 1 m, a ball of mass m = 10 g, flying horizontally, hits absolutely elastically. calculate the momentum of the ball before the impact, if after the impact the thread deflected by an angle a = 60 degrees.

mp = 0.2 kg.

l = 1 m.

mw = 10 g = 0.01 kg.

g = 10 m / s2.

∠α = 60 °.

Rsh -?

The momentum of the ball before the impact is expressed by the formula: Psh = msh * Vsh.

mp * Vp2 / 2 = mp * g * h.

Vп = √ (2 * g * h).

h = l – l * cosα = l * (1 – cosα).

Vп = √ (2 * g * l * (1 – cosα)).

We will assume that the ball stopped at an absolutely elastic impact.

Psh = Pn – the law of conservation of momentum.

Psh = mp * Vp = mp * √ (2 * g * l * (1 – cosα)).

Psh = 0.2 kg * √ (2 * 10 m / s2 * 1 m * (1 – cos60 °)) = 0.63 kg * m / s.

Answer: the momentum of the ball before impact was Psh = 0.63 kg * m / s.