Question

Starting with an initial speed of 3.42 m/s at a height of 0.199 m, a 3.00-kg ball swings downward and strikes a 5.66-kg ball that is at rest, as the drawing shows.

(a) Using the principle of conservation of mechanical energy, find the speed of the 3.00-kg ball just before impact.
(b) Assuming that the collision is elastic, find the velocity (magnitude and direction) of the 3.00-kg ball just after the collision.
(c) Assuming that the collision is elastic, find the velocity (magnitude and direction) of the 5.66-kg ball just after the collision.
(d) How high does the 3.00-kg ball swing after the collision, ignoring air resistance?
(e) How high does the 5.66-kg ball swing after the collision, ignoring air resistance?

Answer 1

The initial speed of the 3.00-kg ball, height, and masses of both balls are given. By applying the principle of conservation of mechanical energy, we can find the speed of the 3.00-kg ball just before impact and the heights of the swings after the collision, ignoring air resistance. Assuming an elastic collision, we can also find the velocities of both balls just after the collision.

Step-by-step explanation:

Conservation of mechanical energy:

(a) To find the speed of the 3.00-kg ball just before impact, we can use the principle of conservation of mechanical energy. Initially, the ball has potential energy given by PE = mgh, where m is the mass (3.00 kg), g is the acceleration due to gravity (9.8 m/s^2), and h is the initial height (0.199 m). At the bottom of the swing, all of the potential energy is converted into kinetic energy, given by KE = (1/2)mv^2, where v is the velocity. Therefore, mgh = (1/2)mv^2, and solving for v gives us the speed just before impact.

(b) Assuming an elastic collision, we can use the law of conservation of momentum to find the velocity of the 3.00-kg ball just after the collision. The initial momentum is given by P = mv, where m and v are the mass and initial velocity, respectively. After the collision, the momentum is still conserved, so we can write m1v1 + m2v2 = m1v1' + m2v2', where m1 and v1 are the mass and initial velocity of the 3.00-kg ball, m2 and v2 are the mass and initial velocity of the 5.66-kg ball, and v1' and v2' are the final velocities. Solving for v1' gives us the velocity of the 3.00-kg ball just after the collision.

(c) The same equation can be used to find the velocity of the 5.66-kg ball just after the collision. Solving for v2' gives us the velocity of the 5.66-kg ball just after the collision.

(d) To find how high the 3.00-kg ball swings after the collision, we can use the conservation of mechanical energy again. At the bottom of the swing, all of the kinetic energy is converted into potential energy, so (1/2)mv^2 = mgh, where h is the final height. Solving for h gives us the height of the swing after the collision.

(e) Similarly, we can use the conservation of mechanical energy to find the height of the 5.66-kg ball swing after the collision.