Block 1 of mass m1 slides from rest along a frictionless ramp from height h=3.1 m and then collides with stationary block 2, which has mass m2 = 4m1. after the collision, block 2 slides into a region where the coefficient of kinetic friction is 0.35 and comes to a stop in distance d with the region. What is the value of distance d if the collision is elastic?

Question

Block 1 of mass m1 slides from rest along a frictionless ramp from height h=3.1 m and then collides with stationary block 2,  which has mass m2 = 4m1. after the collision, block 2 slides into a region where the coefficient of kinetic friction is 0.35 and comes to a stop in distance d with the region. What is the value of distance d if the collision is elastic?

anonymous · Answer

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anonymous · Answer

I've managed to find the final velocity of the 1st block which is 7.7949 m/s by using the conservation of energy which is Ko+uo= Kf + uf

anonymous · Answer

However, I can't seem to determine and understand for the final and initial velocity for the second block. I'm confused, what formula should i use for the elastic collision? Should it be m1v1+m2v2 = m1u1 + m2u2?

irishboy123 · Answer

momentum is always conserved so $mu = mv + 4mw$

here it's *elastic* so you can *also* say that KE is preserved $\frac{1}{2}mu^2 = \frac{1}{2}mv^2 +\frac{1}{2} (4m)w^2$

where w is the velocity of the second block immediately after the collision and $u, v$ are initial and final for the first block

there is also a useful shortcut, using the coefficient of restitution, which allows you to say straight off the bat that $e = 1 =\dfrac{w-v}{u}$, and saves you all that fiddling about with those two previous equations; but if you are only just learning this i'd go through the pain barrier and get used to that first 

you also have $u = \sqrt{2gh}$ but i think you already have that bit

so use your equations to emiliminate v, and get an expression for w.  
🍀

anonymous · Answer

is d = 1.4172 m?

anonymous · Answer

@IrishBoy123

irishboy123 · Answer

from $e = 1 =\dfrac{w-v}{u}$ we have $v = w - u$

so subbing into $mu = mv + 4mw$ we have $u = w - u + 4w$ or $5w = 2u$

and $u = \sqrt{2gh}$, so $w = \dfrac{2}{5} \sqrt{2gh} $

the work done by the friction should equal the kinetic energy of the last block so $ \mu mg x = \frac{1}{2}mw^2$ or $x = \dfrac{m w^2}{2 \mu mg} = \dfrac{ 4(2gh)}{2(25) \mu g} = \dfrac{ 4h}{25 \mu } = 1.4171m$

phew!  we agree....

anonymous · Answer

@IrishBoy123  Thank you for helping me :)

irishboy123 · Answer

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