Abaseball of mass m1 = 0.45 kg is thrown at another ball hanging from the ceiling by a length of string l = 1.35 m. the second ball m2 = 0.61 kg is initially at rest while the baseball has an initial horizontal velocity of v1 = 4.5 m/s. after the collision the first baseball falls straight down (no horizontal velocity.) what is the angle that the string makes with the vertical at the highest point of travel in degrees?

0.947 rad or 54.27 degrees

Explanation:

Suppose the collision is elastic, meaning the momentum is preserved.

Before the collision and the second ball has no momentum

Right after the collision and the first bast ball has no momentum.

Therefore momentum of the first baseball has been transferred to the 2nd ball

After the collision, the second ball would gained kinetic energy, which then would be transferred to potential energy once it reaches its highest point:

By the law of energy conservation:

By law of energy conservation:

So at its highest point, the ball is 0.562 m from the lowest point. Since the ball is hanging on a 1.35 m string, we can calculate the vertical distance from there to the swinging point:

1.35 - 0.562 = 0.788 m

Finally, the angle that string makes with the vertical at the highest point is

what are the statements? x

pythagoras' theorem

square root (-5^2+9^2)

square root (25+81)

square root (106)

angle is tan^-1(9/5)

neg x, pos y