The ratio of the kinetic energy of the block/bullet system immediately after the collision to the initial kinetic energy of the bullet is 0.78 %.
<h3>Final velocity of the block/bullet system</h3>
Apply the principle of conservation of energy to determine the final velocity of the block/bullet system.
K.E = P.E
¹/₂mv² = mgh
¹/₂v² = gh
v² = 2gh
v = √2gh
where;
- h is the maximum height reached by the system
- v is the initial velocity of the system
v = √(2 x 9.8 x 1.1)
v = 4.64 m/s
<h3>Initial velocity of the bullet</h3>
Apply the principle of conservation of linear momentum.
m₁u₁ + m₂u₂ = v(m₁ + m₂)
where;
- u₁ is the initial velocity of the bullet
- u₂ is the initial velocity of the block
- v is the final velocity after collision
- m₁ is mass bullet
- m₂ is mass of block
(0.0075)u₁ + (0.95)(0) = 4.64(0.0075 + 0.95)
0.0075u₁ = 4.4428
u₁ = 4.4428/0.0075
u₁ = 592.37 m/s
<h3>Initial kinetic energy of the bullet</h3>
K.Ei = ¹/₂m₁u₁²
K.Ei = ¹/₂(0.0075)(592.37)²
K.Ei = 1,315.88 J
<h3>Final kinetic energy of the block/bullet system</h3>
K.Ef = ¹/₂(m₁ + m₂)v²
K.Ef = ¹/₂(0.0075 + 0.95)(4.64)²
K.Ef = 10.31 J
<h3>Ratio of final kinetic energy to initial kinetic energy</h3>
= K.Ef/K.Ei x 100%
= (10.31 / 1,315.88) x 100%
= 0.78 %
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Answer: LET'S TAKE GRAVITY ON EARTH AS 10 N/KG AND ON MOON AS 0.6 N/KG
a) weight=15kgx10N/kg
= 150N
b) weight=15x0.6=9N
c) mass=is the same so 15kg
d) mass is measured in kg, weight in NEWTONS OR 
e) when the gravitational force is zero
Explanation:
Answer:
t = 8.3s
Explanation:
V = 100 m/s U = 0 m/s t = ? a = 12 m/s2
V = U + at
100 = 0 + 12 × t = 100 = 12t
12t = 100
t = 100/12 = 25/3
t = 8.3s
Answer:
B. interactive
Explanation:
newtons first law is inertia his second is acceleration and third is interactive.
I will assume that big Joe is big Jim. The equation for the momentum is p=m*v, where m is the mass of the body and v is the velocity. Big Joe has a mass m=105 kg and speed v=5.2 m/s. When we input the numbers:
p=105*5.2=546 kg*(m/s).
So big Joe's momentum before the collision is p=546 kg*(m/s).
