A baseball player uses a bat with mass mbat to hit a ball with mass mball. Right before he hits the ball, the bat's initial velocity is 36 m/s, and the ball's initial velocity is -20 m/s (the positive direction is along the positive x-axis). The bat and ball undergo a one-dimensional elastic collision. Find the speed of the ball after the collision. Assume that mbat is much greater than mball, so the center of mass of the two objects is essentially at the bat.

Question

A baseball player uses a bat with mass mbat to hit a ball with mass mball. Right before he hits the ball, the bat's initial velocity is 36 m/s, and the ball's initial velocity is -20 m/s (the positive direction is along the positive x-axis). The bat and ball undergo a one-dimensional elastic collision. Find the speed of the ball after the collision. Assume that mbat is much greater than mball, so the center of mass of the two objects is essentially at the bat.

Answer 1

u=36 m/s, v(o)= -20 m/s.

Imagine that the bat is at rest, and the ball is approaching to it at the releative velocity
v(x)= v(o)+u =20+36 =56 m/s.
Then in the bat’s frame of reference the ball bounces at the velocity 56 m/s directed in opposite direction. The velocity of the ball in the rigid frame is the sum of the relative velocity of the bounce and velocity of the moving frame of reference.
The projections on the x-axis;
v(x) = 2u(x) –v(ox),
u(x) = u.
v(ox) = -v(o) ,

v(x) = 2u+v(o)

Answer 2

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

1. Conservation of momentum:
Before the collision, the total momentum of the system is given by:
(mbat * vbat) + (mball * vball)

After the collision, the momentum is conserved, so the total momentum of the system is still:
(mbat * vbat') + (mball * vball')

2. Conservation of kinetic energy:
In an elastic collision, the total kinetic energy of the system is conserved. So we have:
(1/2 * mbat * vbat^2) + (1/2 * mball * vball^2) = (1/2 * mbat * vbat'^2) + (1/2 * mball * vball'^2)

Now, let's solve for vball' (the speed of the ball after the collision).

First, let's calculate the velocity of the bat after the collision (vbat') using the conservation of momentum.

Initial momentum = Final momentum
(mbat * vbat) + (mball * vball) = (mbat * vbat') + (mball * vball')

Since mbat is much greater than mball, the velocity of the bat after the collision (vbat') will be negligible compared to vbat.

Therefore, we can ignore the term (mbat * vbat') and reduce the equation to:
(mbat * vbat) + (mball * vball) ≈ mball * vball'

Now let's substitute this equation into the equation for conservation of kinetic energy.

(1/2 * mbat * vbat^2) + (1/2 * mball * vball^2) = (1/2 * mball * vball'^2)

Rearrange the equation:

(mb * vbat^2) + (mball * vball^2) = (mball * vball'^2)

Now, we can solve for vball':

vball'^2 = (mb * vbat^2) + (mball * vball^2) / mball

vball' = sqrt((mb * vbat^2) + (mball * vball^2) / mball)

Substituting the given values:
mbat = mb
vbat = 36 m/s
vball = -20 m/s (negative because it's in the opposite direction)

vball' = sqrt((mb * (36 m/s)^2) + (mball * (-20 m/s)^2) / mball)

Simplifying the equation further, we get:
vball' = sqrt((1296 * mb) + (400 * mball) / mball)

Answer 3

To find the speed of the ball after the collision, we can use the principles of conservation of momentum and conservation of kinetic energy.

1. Conservation of momentum:
In an elastic collision, the total momentum before the collision is equal to the total momentum after the collision. Mathematically, we have:

mbat * vbat_initial + mball * vball_initial = mbat * vbat_final + mball * vball_final

where,
vbat_initial: initial velocity of the bat
vball_initial: initial velocity of the ball
vbat_final: final velocity of the bat
vball_final: final velocity of the ball

In this case, since the center of mass is essentially at the bat and mbat is much greater than mball, the velocity of the bat before and after the collision can be considered nearly the same. Therefore, we can simplify the equation to:

vbat_initial + mball * vball_initial = vbat_final + mball * vball_final

2. Conservation of kinetic energy:
In an elastic collision, the total kinetic energy before the collision is equal to the total kinetic energy after the collision. Mathematically, we have:

(1/2) * mbat * vbat_initial^2 + (1/2) * mball * vball_initial^2 = (1/2) * mbat * vbat_final^2 + (1/2) * mball * vball_final^2

Since the bat's velocity remains constant, we can simplify this equation further:

(1/2) * mball * vball_initial^2 = (1/2) * mball * vball_final^2

Now, we have two equations. We can solve them simultaneously to find the value of vball_final.

First, let's substitute vbat_initial = vbat_final = vbat (approximately) into the momentum conservation equation:

vbat + mball * vball_initial = vbat + mball * vball_final

Simplifying, we get:

mball * vball_initial = mball * vball_final

Now, substitute this equation into the kinetic energy conservation equation:

(1/2) * mball * vball_initial^2 = (1/2) * mball * vball_final^2

Canceling mball and simplifying:

vball_initial^2 = vball_final^2

Taking the square root of both sides:

vball_initial = vball_final

Since the initial velocity of the ball is -20 m/s, we can conclude that the magnitude of the ball's velocity after the collision is 20 m/s.

Therefore, the speed of the ball after the collision is 20 m/s.