Respuesta :
Answer:
Change in momentum of the stone is 3.673 kg.m/s.
Explanation:
Given:
Mass of the ball on the horizontal the surface, m = 0.10 kg
Velocity of the ball with which it hits the stone, v = 20 m/s
According to the question it rebounds with 70% of the initial kinetic energy.
We have to find the change in momentum i.e Δp
Before that:
We have to calculate the rebound velocity with which the object rebounds.
Lets say that the rebound velocity be "v1" and KE remaining after the object rebounds be "KE1".
⇒ [tex]KE_1=0.7\times \frac{mv^2}{2}[/tex]  Â
⇒ [tex]KE_1=0.7\times \frac{0.10\times (20)^2}{2}[/tex]
⇒ [tex]KE_1=0.7\times \frac{0.10\times 400}{2}[/tex]
⇒ [tex]KE_1=14[/tex] Joules (J).
Rebound velocity "v1".
⇒ [tex]KE_1=\frac{m(v_1)^2}{2}[/tex]
⇒ [tex]v_1 = \sqrt{\frac{2KE_1}{m} }[/tex]
⇒ [tex]v_1 = \sqrt{\frac{2\times 14}{0.10} }[/tex]
⇒ [tex]v_1=16.73[/tex]
⇒ [tex]v_1=-16.73[/tex] m/s ...as it rebounds.
Change in momentum Δp.
⇒ [tex]\triangle p= m\triangle v[/tex]
⇒ [tex]\triangle p= 0.10\times (20-(-16.73)[/tex]
⇒ [tex]\triangle p= 0.10\times (20+16.73)[/tex]
⇒ [tex]\triangle p= 0.10\times (36.73)[/tex]
⇒ [tex]\triangle p = 3.673[/tex] Kg.m/s
The magnitude of the change in momentum of the stone is 3.673 kg.m/s.
