A0.500-kg ball traveling horizontally on a frictionless surface approaches a very massive stone at 20.0 m/s perpendicular to wall and rebounds with 70.0% of its initial kinetic energy. what is the magnitude of the change in momentum of the stone?

a) 1.63 kg middot m/s
b) 3.00 kg middot m/s
c) 0.000 kg middot m/s
d) 14.0 kg middot m/s
e) 18.4 kg middot m/s

E)  I = 18.4 N.s

Explanation:

For this exercise let's use momentum momentum

     I = Δp = - p₀

The energy of the stone is only kinetic

    K = ½ m v²

The initial energy is Ko and the final is 70% Ko

      = 0.70 K₀

energy equation

      = 0.7 ½ m v₀²

You can also write

      = ½ m vf²

   ½ m vf² = ½ m (0.7 v₀²)

    = v₀ √ 0.7

Now we can calculate and imposed

     I = m (-vo √0.7) - m vo

     I = m vo (1 +√0.7

     I = 0.5000 20.0 (1.8366)

     I = 18.4 N.s

answer: the correct answer id option(d).

explanation:

snell's law refraction:

the angle of refraction =

the angle of incidence =

refractive index of air = = 1

refractive index of glycerol = 1.47

from the given options, the most closest answer to 63.72° is 65° of option (d).

the bal will not change its velocity.