What is the final velocity of the baseball after being hit by the bat?

AI Thread Summary
The discussion revolves around calculating the final velocity of a baseball after it is hit by a bat, given its initial speed and the work done by the bat. The initial kinetic energy of the baseball is derived from its mass and initial velocity, while the work done on the ball by the bat is also factored in. Participants clarify the use of the work-energy principle, emphasizing that the change in kinetic energy equals the work done. The conversation highlights the importance of correctly applying the conservation of energy equation to find the final velocity. Ultimately, the focus is on resolving the calculations to determine the baseball's speed after impact.
agadag
Messages
59
Reaction score
0

Homework Statement


A pitcher throws a 0.124-kg baseball, and it approaches the bat at a speed of 54.2 m/s. The bat does Wnc = 76.5 J of work on the ball in hitting it. Ignoring the air resistance, determine the speed of the ball after the ball leaves the bat and is 21 m above the point of impact.


Homework Equations



W = E - E0

The Attempt at a Solution


W = E -Eo
76.5=.5(.124)v2-.124(9.8)21
102.0192=.062v2
v = wrong answer!
Please help. I don't have a basics in physics, so please explain from an beginners standpoint.
Thankyou!
 
Physics news on Phys.org
agadag said:

The Attempt at a Solution


W = E -Eo
76.5=.5(.124)v2-.124(9.8)21
102.0192=.062v2
v = wrong answer!

where did you get 9.8 from? It told you 54.2 m/s was the initial velocity. Redo this part. For the second part you need to use projectile motion equations.
 
I used the eqn W=E-E0, since they gave us a non conservative value.
this can b furthere broken down to... 1/2mv2 -mgh. Thats where the 9.8 comes from. As far as projectile motion..how would i use it with no angles?
 
agadag said:
I used the eqn W=E-E0, since they gave us a non conservative value.
this can b furthere broken down to... 1/2mv2 -mgh. Thats where the 9.8 comes from. As far as projectile motion..how would i use it with no angles?

actually as I think about it now, you don't need to use projectile motion.

Initially

all you have is that that change in kinetic energy = work done by that bat.

the initial ke = 1/2m(54)2.

For the second part you need to use the part in red
 
Im confused. I got KE = 180.792. however, I am trying to find final velocity of the ball. i initially tried the eqn in red and it did not work for me. Can you please demonstrate?
 
agadag said:
Im confused. I got KE = 180.792. however, I am trying to find final velocity of the ball. i initially tried the eqn in red and it did not work for me. Can you please demonstrate?

Conservation of energy right

Initially it has ke given by 1/2mu2. This energy is converted into kinetic energy (it is moving with a different velocity v) and work done by the bat.

so applying the law of conservation of energy we will get 1/2mu2=1/2mv2+Wbat.

So can you find 'v' given Wbat=76.5J and u=54.2m/s?
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top