(PLS HELP 20 POINTS, IM TOO DUMB FOR THIS) What is the total momentum of the system after the collision?

Physics

1 answer:

Dmitry_Shevchenko [17]3 years ago

6 0

Using the law of conservation of momentum
m1u1+m2u2=m1v1+m2v2
Where m1 is mass of first object
m2 is mass of second object
u1 and u2 are initial velocities of object 1 and 2 respectively
v1 and v2 are final velocities of object 1 and 2 respectively
Here, they are moving as a system after collision. Thus they will posses same final velocity
m1u1 +m2u2=v(m1+m2)
Substituting values
600*4+0=v(600+400)
2400=v*1000
v=2.4 m/s

Now momentum of system
p=Mv
p=(600+400)*2.4
p=1000*2.4
Therefore p=2400 kg m/s
Hope this helps :)

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$\begin{aligned}\frac{d}{d q_{1}}[F] &= \frac{d}{d q_{1}} \left[\frac{k}{r^{2}}\, (q\, q_{1} - {q_{1}}^{2})\right] \\ &= \frac{k}{r^{2}}\, \left[\frac{d}{d q_{1}} [q\, q_{1}] - \frac{d}{d q_{1}}[{q_{1}}^{2}]\right]\\ &= \frac{k}{r^{2}}\, (q - 2\, q_{1})\end{aligned}$ .

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$\begin{aligned}\frac{d^{2}}{{d q_{1}}^{2}}[F] &= \frac{d}{d q_{1}} \left[\frac{k}{r^{2}}\, (q - 2\, q_{1})\right] \\ &= \frac{(-2)\, k}{r^{2}}\end{aligned}$ .

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