Answer:
- Here we use the conservation of momentum theorem.
- m stands for mass, and v stands for velocity. The numbers refer to the respective objects.
- m1v1 + m2v2 = m1vf1 + m2vf2
- Since the equation is perfectly inelastic, the final velocity of both masses is the same. Let’s account for this in our formula.
- m1v1 + m2v2 = vf(m1 + m2)
<u>Let’s substitute in our givens.</u>
(0.002 kg)(700 m/s) + (5 kg)(0 m/s) = vf(0.002 kg + 5 kg)
I assume you are proficient in algebra I, so I will not include the steps to simplify this equation.
Note that I have considered the bullet’s velocity to be in the positive direction,
The answer is vf = 0.280 m/s
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Potential energy is measured using formula Ep=mgh
m=mass (kg)
g= acceleration due to gravity (which is 9.8 on earth)
h= height in metres above ground
For this question
m=0.1
g=9.8
h=1
So Ep=0.1(9.8)(1)
Ep=0.98 Joules
When it is dropped all of this potential energy is converted into kinetic energy which can be measured using formula
Ek=1/2m(v^2) (v=final velocity)
Since all potential energy in this q is converted to kinetic we know Ek=0.98Joules and our mass is the same (0.1kg)
So when we sub everything in we get
0.98=1/2(0.1)(v^2)
0.98=0.05(v^2)||divide both side by 0.05
19.6=v^2 ||square root both sides
v=4.4 m/s
Answer:
In vector form:
S=30j^+20i^+302cos45o(−i^)+302sin45o(−j^)
=30j^+20i^−30i^−30j^
=10i^
Explanation:
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