A golf ball, mass 0.048 kg, rests on a tee. It is struck by a golf club with an effective mass of 0.225 kg and a speed of 44 m/s. Assuming the collision is elastic, find the speed of the ball when it leaves the tee (remember: you can not assume the club is at rest after the collision).
Well, well, well, looks like the golf ball is in for a ride! Let's tackle this problem with a little humor, shall we?
First, we need to understand that during an elastic collision, momentum is conserved. So, we can use the equation:
m1v1 + m2v2 = m1v1' + m2v2'
Where:
m1 = mass of the ball
v1 = initial velocity of the ball
m2 = mass of the club
v2 = initial velocity of the club
v1' = final velocity of the ball
v2' = final velocity of the club
Now, let's plug in the given values. We have:
m1 = 0.048 kg
v1 = unknown (that's what we're trying to find!)
m2 = 0.225 kg
v2 = -44 m/s (negative because the club moves in the opposite direction)
v1' = unknown (let's call it V)
v2' = unknown (let's call it U)
Using the conservation of momentum equation, we have:
(0.048 kg)(v1) + (0.225 kg)(-44 m/s) = (0.048 kg)(V) + (0.225 kg)(U)
Now, let's use a little algebra magic. We need to find the final velocities V and U, so let me wave my wand and rearrange the equation:
(0.048 kg)(v1) + (0.225 kg)(-44 m/s) = (0.048 kg)(V) + (0.225 kg)(U)
(0.048 kg)(v1) - (0.048 kg)(V) = (0.225 kg)(U) + (0.225 kg)(44 m/s)
(0.048 kg)(v1 - V) = (0.225 kg)(U) + (0.225 kg)(44 m/s)
v1 - V = (0.225 kg)(U)/0.048 kg + (0.225 kg)(44 m/s)/0.048 kg
v1 - V = (4.6875 kg)(U) + (9.1667 kg m/s)
Now, we're almost there! Let's solve for V:
V = v1 - (4.6875 kg)(U) - (9.1667 kg m/s)
And there you have it, the speed of the ball when it leaves the tee. Just substitute the appropriate values and do a little calculation, and you'll have your answer. Swing away, my friend!
To find the speed of the ball when it leaves the tee after the collision, we can use the conservation of momentum principle.
According to the principle of conservation of momentum, the total momentum before the collision is equal to the total momentum after the collision.
Let's denote the initial velocity of the golf ball as v1, the final velocity of the golf ball as v2, and the initial velocity of the golf club as u.
The total momentum before the collision is the sum of the momenta of the ball and the club, while the total momentum after the collision is the momentum of the ball alone.
The momentum of an object can be calculated by multiplying its mass by its velocity:
Momentum = mass x velocity
Before the collision:
Total initial momentum = Momentum of the ball + Momentum of the club
= (mass of the ball * velocity of the ball) + (mass of the club * velocity of the club)
= (0.048 kg * v1) + (0.225 kg * u)
After the collision:
Final momentum of the ball = mass of the ball * final velocity of the ball
= 0.048 kg * v2
According to the conservation of momentum principle, the total initial momentum is equal to the final momentum of the ball:
(0.048 kg * v1) + (0.225 kg * u) = 0.048 kg * v2
Now, we can substitute the given values and solve for v2:
(0.048 kg * v1) + (0.225 kg * 44 m/s) = 0.048 kg * v2
0.048 kg * v1 + 9.9 kg.m/s = 0.048 kg * v2
Rearranging the equation, we can isolate v2:
0.048 kg * v2 = (0.048 kg * v1) + 9.9 kg.m/s
v2 = ((0.048 kg * v1) + 9.9 kg.m/s) / 0.048 kg
Now you just need to substitute the given values for v1 and solve for v2 to find the speed of the ball when it leaves the tee after the collision.
To find the speed of the ball when it leaves the tee after being struck by the golf club, we can use the principles of conservation of momentum and kinetic energy.
1. First, let's calculate the initial momentum of the system. The initial momentum (p_i) of the system consisting of the golf ball and the golf club before the collision can be calculated using the formula:
p_i = m_club * v_club + m_ball * v_ball
where:
- m_club is the mass of the golf club (0.225 kg)
- v_club is the speed of the golf club (44 m/s)
- m_ball is the mass of the golf ball (0.048 kg)
- v_ball is the initial speed of the golf ball (which we want to find)
So, p_i = (0.225 kg) * (44 m/s) + (0.048 kg) * v_ball
2. Next, let's calculate the final momentum (p_f) of the system after the collision. According to the principle of conservation of momentum, the total momentum of a system before an event (in this case, the collision) is equal to the total momentum of the system after the event. Therefore:
p_f = m_club * v'_club + m_ball * v'_ball
where:
- v'_club is the final velocity of the golf club after the collision (unknown)
- v'_ball is the final velocity of the golf ball after the collision (which we want to find)
3. Since the collision is assumed to be elastic, the kinetic energy of the system is conserved. The initial kinetic energy (KE_i) of the system is given by:
KE_i = (1/2) * m_club * v_club^2 + (1/2) * m_ball * v_ball^2
And the final kinetic energy (KE_f) of the system is given by:
KE_f = (1/2) * m_club * v'_club^2 + (1/2) * m_ball * v'_ball^2
From the conservation of kinetic energy, we have KE_i = KE_f.
Now, we can proceed to solve for v_ball by equating the initial and final momenta, and using the kinetic energy conservation equation:
m_club * v_club + m_ball * v_ball = m_club * v'_club + m_ball * v'_ball
(1/2) * m_club * v_club^2 + (1/2) * m_ball * v_ball^2 = (1/2) * m_club * v'_club^2 + (1/2) * m_ball * v'_ball^2
Substituting the given values into the equations and solving the resulting system of equations will allow us to find the final velocity of the ball (v'_ball).
Because this is an elastic collision, you will need 2 equations. I'm sure you will be familiar with these:
m1v1 + m2v2 = m1v1' + m2v2'
v1 - v2 = v2' - v1'
Let's make it that...
m1 = mass of golf ball
v1 = initial velocity of golf ball
v1' = final velocity of golf ball
m2 = mass of golf club
v2 = initial velocity of golf club
v2' = final velocity of golf club
Okay, now that is settled, plug in what you know.
(0.048 kg)(0 m/s) + (0.225 kg)(44 m/s) = (0.048 kg)v1' + (0.225 kg)v2'
(0 m/s) - (44 m/s) = v2' - v1'
I'll leave the algebra to you. You should get that the velocity of the golf ball (v1') is 72.5 m/s.