Question

Person A with a mass of 70kg has an initial velocity of 5 m/s. Is it possible for a second person to collide elastically with person A and have both their final velocities be 0? If so, what does the mass/initial velocity of the second person have to be? If not, why not? A large wheel is rolling along at a linear velocity of 5 m/s. The moment of inertia for the wheel is (3/5)MR2. It rolls up an inclined piane (angle of 40 degrees). What is the maximum height that the wheel reaches? 6) 7)

Answer 1

(6) For elastic collision –

(a) Conservation of momentum shall be applied

(b) Conservation of energy shall be applied.

Suppose the mass of the second person = m2

And its initial velocity = v2

Suppose the final velocity of the two persons after the collision is v1’ and v2’.

Apply conservation of momentum –

70*5 kg*m/s + m2*v2 = 70*v1’ + m2*v2’

=> 350 + m2*v2 = 70*v1’ + m2*v2’ -------------------------------------(i)

Looking at the equation (i), we find that at any case, it is not possible that value of v1’ = 0 and v2’ = 0

So, we can say that in the case of elastic collision, final velocity of both the persons equal to ZERO is not possible.

(7) Total kinetic energy of the wheel –

KE(total) = (1/2)*M*v^2 + (1/2)*I*w^2

Where, I = Moment of Inertia of the wheel = (3/5)*M*R^2

w = angular velocity of the wheel = v/R

So –

KE(total) = (1/2)*M*v^2 + (1/2)* (3/5)*M*R^2*(v/R)^2

                = (1/2)*M*v^2 + (3/10)*M*v^2

                = 0.80*M*v^2

Now suppose, h meter is the height up to which the wheel climbs up.

So, potential energy attained by the wheel, PE = M*g*h

According to conservation of energy –

PE = KE(total)

=> M*g*h = 0.80*M*v^2

Cancel ‘M’ from both sides –

h = (0.80*v^2) / g

Put the values-

h = (0.80*5.0^2) / 9.81 = 2.04 meter (Answer)