What is the speed of the 100g ball in Newton's Cradle after being lifted 3.0cm?

  • Thread starter Thread starter madeeeeee
  • Start date Start date
AI Thread Summary
The discussion centers on calculating the speed of a 100 g ball in a Newton's Cradle after being lifted 3.0 cm. The potential energy (Eg) is calculated using the formula Eg = mgh, resulting in 0.0294 J. The kinetic energy (Ek) is expressed as Ek = 1/2mv^2, leading to the equation 0.0294 J = 1/2(0.1 kg)(v)^2. Participants discuss the application of the conservation of energy principle, suggesting that total energy at the top (Eg) equals kinetic energy at the bottom (Ek). The final equation proposed is mgh = 1/2mv^2, indicating a correct approach to solving for the speed.
madeeeeee
Messages
85
Reaction score
0

Homework Statement


In David and Mervin's Newton's Cradle the 100 g ball was lifted a height of 3.0 cm. With what speed would it collide with the other balls at the bottom of its motion?


Homework Equations


Eg=mgh
Ek=1/2mv^2
100g=0.1 kg
3.0cm=0.03m


3. The Attempt at a Solution [/b]
Eg=mgh
Eg=(0.1kg)(9.8m/s^2)(0.03m)
Eg=0.0294J
Ek=1/2mv^2
0.0294J=1/2(0.1kg)(v)^2

Am i on the right track? I think i am doing something wrong
 
Physics news on Phys.org
What do you think you're doing wrong? Or why do you think something is wrong?
 
ok so what if i use the conservation of energy equation
Et=Ek=Eg

would the final equation be
mgh = 1/2mv2^2
 
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