A regulation 145 g baseball can be hit at speeds of 100 mph. If a line drive is hit essentially horizontally at this speed and is caught by a 65 kg player who has leapt directly upward into the air, what horizontal speed (in cm/s) does he acquire by catching the ball?
To find the horizontal speed acquired by the player after catching the ball, we can use the principle of conservation of momentum. According to this principle, the total momentum before and after an event remains constant, assuming no external forces are acting.
Let's assume the initial horizontal speed of the baseball is Vb and the final horizontal speed of the player is Vp.
The momentum of the baseball before catching it can be calculated as:
Momentum before = mass of the baseball * initial horizontal speed
= 145 g * Vb
The momentum of the player after catching the ball can be calculated as:
Momentum after = mass of the player * final horizontal speed
= 65 kg * Vp
Since the total momentum is conserved, the momentum before catching the ball should be equal to the momentum after catching it:
145 g * Vb = 65 kg * Vp
To find the horizontal speed in cm/s, we need to convert the units:
145 g * Vb = 65 kg * Vp
Converting 145 g to kg:
0.145 kg * Vb = 65 kg * Vp
Now we can solve for Vp:
Vp = (0.145 kg * Vb) / 65 kg
Simplifying this equation gives us the formula for Vp:
Vp = (0.00223 * Vb) cm/s
Therefore, the player acquires a horizontal speed of (0.00223 * Vb) cm/s by catching the ball.
To determine the horizontal speed acquired by the player when catching the baseball, we need to apply the principle of conservation of momentum.
1. Calculate the initial momentum of the baseball:
Momentum (p) = mass (m) * velocity (v)
The mass of the baseball is given as 145 g, which is 0.145 kg.
The velocity of the baseball is given as 100 mph, which needs to be converted to m/s.
1 mph = 0.44704 m/s
So, the velocity of the baseball is 100 mph * 0.44704 m/s = 44.704 m/s.
Therefore, the initial momentum of the baseball is: p(baseball) = 0.145 kg * 44.704 m/s.
2. Calculate the final momentum of the player and the baseball system:
Since the player catches the baseball while leaping straight up, the vertical component of momentum becomes zero. Therefore, we only need to consider the horizontal component of momentum.
The mass of the player is given as 65 kg.
Let's assume the horizontal speed acquired by the player upon catching the ball is v(player).
The final momentum of the player and baseball system is then given by:
p(final) = (0.145 kg * 44.704 m/s) + (65 kg * v(player))
3. Apply the conservation of momentum principle:
According to the conservation of momentum, the initial momentum and final momentum of a system must be equal.
Equating the initial momentum of the baseball to the final momentum of the player and baseball system:
p(baseball) = p(final)
0.145 kg * 44.704 m/s = (0.145 kg * 44.704 m/s) + (65 kg * v(player))
Simplifying this equation, we get:
0 = 65 kg * v(player)
4. Solve for the horizontal speed acquired by the player:
Since 65 kg * v(player) = 0, this means the horizontal speed acquired by the player upon catching the ball is zero. Therefore, the player does not acquire any horizontal speed.
In conclusion, the player does not acquire any horizontal speed (0 cm/s) by catching the baseball.