Inelastic collision & friction

AI Thread Summary
A 1 kg block connected to a spring collides with a 2 kg block moving at 6 m/s, resulting in both blocks sticking together and moving at a final velocity of 4 m/s after the collision. The discussion then shifts to finding the coefficient of kinetic friction on a rough surface, given that the spring compresses by 0.2 m. The initial calculations led to an incorrect coefficient greater than 1, prompting a reevaluation of energy conservation principles. It was clarified that the final velocity should not be assumed to be zero during spring compression, as the blocks are still in motion. Ultimately, the correct coefficient of kinetic friction was determined to be 0.6.
Hawraa
Messages
5
Reaction score
0
the question is :
a 1 kg block at rest on a horizontal surface is connected to an unstreched spring with a spring constant k=1000 N/m . A 2 kg block whose speed is 6 m/s collides with 1kg block. if the two blocks stick together and move on a rough surface after collision: find
1- the final velocity just after collision in ( m/s ) ?

i found this using momentum equation
m1v1=m2v2
(2)(6) = (2+1)v2
v2( which is the common velocity that the 2 masses are going to move with )= 12/3=4m/s

now the second part
find the coefficient of the kinetic friction of the rough surface , if the spring is compressed by maximum distance of x=0.2 m ?

i tried to solve it using this equation but finally i get a coefficient which is bigger than 1 !

delta K + delta U = work done by friction
{(1/2)(3)(4)square - (1/2)(2)(6)square } + { (1/2)(1000)(0.2)square - (1/2)(1000)(zero)square} = work done by friction

wf = 8
- kinetic friction coefficient . Normal force . d = 8
- kinetic friction coefficient . (3)(10) . (0.2) = 8
i get this answer kinetic friction coefficient = -1.3
canu pleez tell me where did i go wrong in this ?
 
Physics news on Phys.org
Hawraa said:
delta K + delta U = work done by friction
{(1/2)(3)(4)square - (1/2)(2)(6)square } + { (1/2)(1000)(0.2)square - (1/2)(1000)(zero)square} = work done by friction
Not sure what you're doing here. The starting point is immediately after the collision--the 6 m/s speed is no longer relevant. Compare the initial and final energy of the 2-mass system.
 
so i consider the final velocity is zero ?? as the spring compresses to the maximum distance ?
 
Hawraa said:
so i consider the final velocity is zero ?? as the spring compresses to the maximum distance ?
Sure. As long as it's still moving, it hasn't reached full compression. Right?
 
thank u i got it 0.6 i think it's right :D
 
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...
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