Two particles approach each other with equal and opposite speed v. the mass of one particle is m, and the mass of the other particle is nm, where n is just a unitless number. snapshots of the system before, during, and after the elastic collision are shown above. after the collision the first particle moves in the exact opposite direction with speed 2.95v, and the speed of the second particle is unknown. what is the value of n?

The value of n = 79

Momentum is the product of the mass of an object and its velocity

If 2 objects with mass A and mass B collide with velocity va and vb, then the law of conservation of momentum will occur:

can be stated:

the total momentum before the collision is equal to the total momentum after the collision (in the absence of external forces)

Collisions between these 2 objects can be

elastic, inelastic, and completely inelastic (the colliding objects stick together)

For elastic collision applies

Two particles approach each other with equal and opposite speed v. , because the velocity is a vector, then the velocity of objects a and objects b is the same but has a different sign (according to the direction, we consider the right direction positive)

va = v ( to the right)

vb = -v (to the left)

after an elastic collision occurs

va '= -2.95 v ( to the left)

vb '= v' (to the right)

so equation 2 becomes

va'-vb '= vb - va

-2.95v - vb '= -v - v

-2.95v -vb '= -2v

-vb '= - 2v + 2.95v

-vb '= 0.95v

vb '= -0.95v

We input in equation 1 (ma = m, mb = nm)

mava + mbvb = mava '+ mbvb'

m.v-nmv = m.-2.95v + nm. -0.95v -> we divide both segments by m

v-nv = -2.95v-n.0.95v (combine like terms)

v + 2.95v = nv-0.95v

3.95v = n0.05v

n = 3.95v: 0.05v

n = 79

law of motion  

link  

displacement of A skateboarder

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The distance of the elevator  

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divide the right side by the left side

0.024÷1.2=0.02 for example