In collision that are categorized as elastic, the total kinetic energy of the system is preserved such that,
KE1 = KE2
The kinetic energy of the system before the collision is solved below.
KE1 = (0.5)(25)(20)² + (0.5)(10g)(15)²
KE1 = 6125 g cm²/s²
This value should also be equal to KE2, which can be calculated using the conditions after the collision.
KE2 = 6125 g cm²/s² = (0.5)(10)(22.1)² + (0.5)(25)(x²)
The value of x from the equation is 17.16 cm/s.
Hence, the answer is 17.16 cm/s.
Friction force is when you rub 2 things together and they get warm. Motion, on the other hand, is if your walking along the sidewalk - you hardly get warmer -------
Unless it's a colder day outside and you're walking SO you decide to rub your hands together to get warm, but if you were just walking , its motion and only motion - no friction :):)
The momentum of ball is given by:
Since both have the same momentum, we have:
Number 3If you notice any mistake in my english, please let me know, because i am not native.
Answer:
v = √2G
/ R
Explanation:
For this problem we use energy conservation, the energy initiated is potential and kinetic and the final energy is only potential (infinite r)
Eo = K + U = ½ m1 v² - G m1 m2 / r1
Ef = - G m1 m2 / r2
When the body is at a distance R> Re, for the furthest point (r2) let's call it Rinf
Eo = Ef
½ m1v² - G m1
/ R = - G m1
/ R
v² = 2G
(1 / R - 1 / Rinf)
If we do Rinf = infinity 1 / Rinf = 0
v = √2G
/ R
Ef = = - G m1 m2 / R
The mechanical energy is conserved
Em = -G m1
/ R
Em = - G m1
/ R
R = int ⇒ Em = 0