How Fast Can a Car Hit a Spring Bumper Without Damage?

AI Thread Summary
The discussion revolves around calculating the maximum velocity a 1000kg car can hit a spring bumper with a spring constant of 8 x 10^5 N/m without causing damage. The initial calculations yielded an incorrect maximum velocity of 4.2 m/s, while the expected answer was 34 m/s. Participants debated the reasonableness of the 34 m/s figure, ultimately concluding that it seemed excessive. The conversation highlighted the importance of verifying the accuracy of textbook answers. The final consensus suggested that the lower speed of 4.2 m/s felt more plausible for avoiding damage.
Smartgurl
Messages
15
Reaction score
0

Homework Statement


A small truck is equipped with a rear bumper that has a spring constant of 8 x 10^5 N/m. The bumper can be compressed 15 cm without causing damage to the truck. What is the maximum velocity with which a solid 1000kg car can collide with the bumper without causing damage to the truck?[/B]

Homework Equations


Ek = 1/2 mv^2
El = 1/2 kx^2

The Attempt at a Solution


I made these two equations equal to each other and got the answer 4.2 m/s. The right answer is 34 m/s. I'm way off and not sure what I'm doing wrong
 
Physics news on Phys.org
Smartgurl said:

Homework Statement


A small truck is equipped with a rear bumper that has a spring constant of 8 x 10^5 N/m. The bumper can be compressed 15 cm without causing damage to the truck. What is the maximum velocity with which a solid 1000kg car can collide with the bumper without causing damage to the truck?[/B]

Homework Equations


Ek = 1/2 mv^2
El = 1/2 kx^2

The Attempt at a Solution


I made these two equations equal to each other and got the answer 4.2 m/s. The right answer is 34 m/s. I'm way off and not sure what I'm doing wrong
Does 34 m/s seem like a reasonable speed at which damage would be avoided?

Chet
 
Chestermiller said:
Does 34 m/s seem like a reasonable speed at which damage would be avoided?

Chet
No now that I think about it that's not reasonable. Does that mean I'm actually doing it right?
 
Smartgurl said:
No now that I think about it that's not reasonable. Does that mean I'm actually doing it right?
4.2 m/s sounds much more like it, doesn't it?

Chet
 
Chestermiller said:
4.2 m/s sounds much more like it, doesn't it?

Chet
Yes it does! I guess it's one of those rare occasions when the textbook is wrong! Thanks! I'll definitely make sure the answer is reasonable next time
 
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}...
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...

Similar threads

Back
Top