Two balls of equal mass collide and stick together as shown in the figure. The initial velocity of ball B is twice that of ball A. (Take è = 34°.)


(a) Calculate the angle above the horizontal of the motion of mass A + B after the collision.
1 Units are required for this answer. 12.7 degree

(b) What is the ratio of the final velocity of the mass A + B to the initial velocity of ball A, vf/vA?
2
cannot_evaluate 1.27

(c) What is the ratio of the final energy of the system to the initial energy of the system, Ef/Ei?
3
cannot_evaluate 0.65

Is the collision elastic or inelastic?

4elastic
inelastic

Sapo

student

universiity

To solve this problem, we can use the principles of conservation of momentum and conservation of kinetic energy.

Let's start with part (a) - finding the angle above the horizontal of the motion of mass A + B after the collision. We can use the conservation of momentum to solve this.

The principle of conservation of momentum states that the total momentum before the collision is equal to the total momentum after the collision. In this case, the initial momentum of ball B is twice that of ball A.

Let's represent the initial velocity of ball A as vA and the initial velocity of ball B as vB. Since ball B has twice the initial velocity of A, we have vB = 2vA.

After the collision, the two balls stick together and move as one mass A + B. Let's represent the final velocity of this combined mass as vf.

Using the conservation of momentum, we can write the equation as:
(mass A)(vA) + (mass B)(vB) = (mass A + mass B)(vf)

Since the two balls have equal mass, we can simplify the equation to:
2(mass A)(vA) = (2mass A)(vf)

Simplifying further, we have:
2vA = vf

Now, we need to find the angle above the horizontal of the motion of mass A + B. We can use trigonometry to do this.

The initial velocity of ball A is vA at an angle of 34° above the horizontal. Since the final velocity of mass A + B is vf and we know that 2vA = vf, we can find the angle using the formula:

angle = arctan(vf/vA)
= arctan(2vA/vA)
= arctan(2)
≈ 63.4°

Therefore, the angle above the horizontal of the motion of mass A + B after the collision is approximately 63.4°.

Moving on to part (b), we need to find the ratio of the final velocity of the mass A + B to the initial velocity of ball A, vf/vA.

From our previous calculations, we know that vf = 2vA. Therefore, the ratio can be written as:
vf/vA = (2vA)/vA = 2

So, the ratio is 2.

For part (c), we need to find the ratio of the final energy of the system to the initial energy of the system, Ef/Ei.

The principle of conservation of kinetic energy states that the total kinetic energy before the collision is equal to the total kinetic energy after the collision.

The initial kinetic energy of the system is given by:
Ei = (1/2)(mass A)(vA^2) + (1/2)(mass B)(vB^2)

The final kinetic energy of the system is given by:
Ef = (1/2)(mass A + mass B)(vf^2)

Substituting the values we found earlier:
Ef = (1/2)(2mass A)((2vA)^2)
= (1/2)(4mass A)(vA^2)

Now, we can simplify the equation to find the ratio of Ef to Ei:
Ef/Ei = [(1/2)(4mass A)(vA^2)] / [(1/2)(mass A)(vA^2) + (1/2)(mass B)(vB^2)]

Simplifying further, we have:
Ef/Ei = 4(mass A)(vA^2) / [(mass A)(vA^2) + (mass B)(vB^2)]

Since mass A = mass B and vB = 2vA, we have:
Ef/Ei = 4vA^2 / (vA^2 + (2vA)^2)
= 4 / 5
≈ 0.8

Therefore, the ratio of the final energy to the initial energy is approximately 0.8.

Finally, based on the information given in the problem, the collision is inelastic. This is because the two balls collide and stick together, resulting in a loss of kinetic energy. In an elastic collision, kinetic energy would be conserved, but in this case, we see a decrease in energy, causing the collision to be inelastic.