Starting with an initial speed of 3.86 m/s at a height of 0.480 m, a 1.78-kg ball swings downward and strikes a 4.75-kg ball that is at rest, as the drawing shows. (a) Using the principle of conservation of mechanical energy, find the speed of the 1.78-kg ball just before impact. (b) Assuming that the collision is elastic, find the velocity (magnitude and direction) of the 1.78-kg ball just after the collision. (c) Assuming that the collision is elastic, find the velocity (magnitude and direction) of the 4.75-kg ball just after the collision. (d) How high does the 1.78-kg ball swing after the collision, ignoring air resistance? (e) How high does the 4.75-kg ball swing after the collision, ignoring air resistance?
To answer these questions, we will use the principle of conservation of mechanical energy, which states that the total mechanical energy of a system remains constant if there are no external forces acting on it.
(a) To find the speed of the 1.78-kg ball just before impact, we can use the conservation of mechanical energy. The mechanical energy is given by the sum of kinetic energy (KE) and potential energy (PE).
Initially, the ball has kinetic energy and potential energy given by:
KE1 = (1/2) * mass1 * velocity1^2
PE1 = mass1 * g * height1
where mass1 = 1.78 kg, velocity1 = 3.86 m/s, g is the acceleration due to gravity (approximately 9.8 m/s^2), and height1 = 0.480 m.
The total mechanical energy before the collision is:
E1 = KE1 + PE1
Since there are no external forces acting on the system, the total mechanical energy is conserved.
At impact, the 1.78-kg ball collides with the 4.75-kg ball. Therefore, the total mechanical energy after the collision is the same as before.
Let's denote the speed of the 1.78-kg ball just after the collision as v2. We can now write the equation for the total mechanical energy after the collision:
E2 = (1/2) * mass1 * v2^2 + mass1 * g * height2 + (1/2) * mass2 * v3^2 + mass2 * g * height3
where mass2 = 4.75 kg and v3 is the speed of the 4.75-kg ball just after the collision.
(b) Assuming that the collision is elastic, we know that the total mechanical energy is conserved. Therefore, we can equate E1 and E2:
(1/2) * mass1 * velocity1^2 + mass1 * g * height1 = (1/2) * mass1 * v2^2 + mass1 * g * height2 + (1/2) * mass2 * v3^2 + mass2 * g * height3
Substituting the given values and solving this equation will give us the speed of the 1.78-kg ball just after the collision (v2).
(c) Similarly, if the collision is elastic, we can find the velocity of the 4.75-kg ball just after the collision (v3). By using the equation derived in part (b) and solving for v3, we can determine both the magnitude and direction of the velocity.
(d) To find how high the 1.78-kg ball swings after the collision, we need to calculate the new height (height2) for the 1.78-kg ball. This can be done by using the conservation of mechanical energy again:
E2 = (1/2) * mass1 * v2^2 + mass1 * g * height2 + (1/2) * mass2 * v3^2 + mass2 * g * height3
Substituting the known values and solving for height2 will give us the answer.
(e) Similarly, to determine how high the 4.75-kg ball swings after the collision, we need to calculate the new height (height3) for the 4.75-kg ball. Again, we can use the conservation of mechanical energy equation obtained in part (b) and solve for height3.
Please note that to find the values of velocities and heights, you will need to substitute the given values into the derived equations and solve them using algebraic manipulation.