Answer:
PE=mgh
KE=<u>1</u> mv²
2
hope its helpful for u ..
IF the bag is suspended freely in mid-air and free to move
without friction, then the linear momentum is the same both
before and after the hit.
Momentum of the bullet before the hit = ( m₁ v₁ ).
Momentum of the (bullet + bag) after the hit = (m₁ + m₂) (v₂) .
Momentum is conserved, so we can say that ...
( m₁ v₁ ) = (m₁ + m₂) (v₂)
v₂ = ( m₁ v₁ ) / (m₁ + m₂)
Usually the mass of the bag is much much more than the mass
of the bullet, so the mass of the bullet can be ignored after the
hit, and the formula is very closely ...
v₂ = v₁ ( m₁ / m₂ ) .
Stop apologizing for your English. It's better than many native
Americans writing on this site. Be proud.
Answer:
Change in momentum, 
Explanation:
It is given that,
Mass of the basketball, m = 601 g = 0.601 kg
The basketball makes an angle of 29 degrees to the vertical, it hits the floor with a speed, v = 6 m/s
It bounces up with the same speed, again moving to the right at an angle of 29 degree to the vertical. We need to find the change in momentum. It is given by :
So, the change in momentum of the basketball is 6.3 kg-m/s. Hence, this is the required solution.
Displacement = (straight-line distance between the start point and end point) .
Since the road east is perpendicular to the road north,
the car drove two legs of a right triangle, and the magnitude
of its final displacement is the hypotenuse of the triangle.
Length of the hypotenuse = √ (215² + 45²)
= √ (46,225 + 2,025)
= √ 48,250
= 219.7 miles .
