Question

A snowball strikes and sticks to a realty yard sign that is initially at rest, making the sign swing upward. How does the total kinetic energy of the snowball and sign just after the collision compare with the kinetic energy of the ball just before?

Answer 1

Answer:

Some of the snowball's initial kinetic energy is lost in form of frictional, heating or due to deformation of bodies, once inelastic collision has happened.

Hence, the kinetic energy of the snowball and sign after collision is less than the kinetic energy of the snowball before collision.

Explanation:

This collision, where the two bodies that collide and they stick together after collision is called inelastic collision.

The one where the two bodies don't stick together after collision is called elastic collision.

In elastic collision, both momentum and kinetic energy are conserved with the sum of each of these quantities before collision being equal to the sum after collision.

But, when it comes to inelastic collision, only momentum is conserved.

The kinetic energy isn't conserved as part of the initial kinetic energy is lost in some form to ensure that the the kinetic energy of the snowball plus sign system after collision is smaller compared to the kinetic energy of the snowball before collision.

The forms in which the kinetic energy is lost include through heating effects arising from a vibration of the atoms of the two bodies involved, through the deformation of either or both bodies etc.

Hope this Helps!!!

Answer 2

Answer:

$\frac{E_a}{E_b}=\frac{m_s+m_y}{m_s}\frac{v^2}{v_s^2}$

Explanation:

In this case we have an inelastic collision. However, we can consider that the kinetic energy (of the snowball) before the collision must equal the total kinetic energy of (snowball and yard sign) after the collision:

$E_b=E_a\\\\\frac{1}{2}m_sv_s^2=\frac{1}{2}(m_s+m_y)v^2$

where ms is the mass of the snowball, my is the mass of the yard sign, vs is the velocity of the snowball before the collision and v is the velocity of the snowball joint the yard sign.

By comparing both total energies Eb and Ea (before and after) we obtain:

$\frac{E_a}{E_b}=\frac{m_s+m_y}{m_s}\frac{v^2}{v_s^2}$

hope this helps!!