Respuesta :
Answer:2mv
Explanation:
Given
Ball travelling with speed v hits the wall and rebound with same speed
Initial momentum of ball is [tex]P_i=m\cdot v[/tex]
Momentum after the collision is [tex]P_f=m\cdot (-v )[/tex]
Change in momentum is [tex]\Delta P=P_i-P_f[/tex]
[tex]\Delta P=mv-(-mv)[/tex]
[tex]\Delta P=2mv[/tex]
Therefore magnitude of change in momentum is [tex]2 mv[/tex]
Answer: (MbxUb) + MbVb)= M( V+U). Where U=0, initial velocity.
It can be express as M1V1=M2V2.
Meaning momentum before collision is equal moment after collision.
Explanation:
If the ball returns with the same speed, definitely there will be no loss in the kinetic energy, so it returns with the same momentum before the impact.
