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svlad2 [7]
3 years ago
10

Two blocks with a mass of 2 kg collide in an elastic collision. The initial velocity of Block 1 is 10 m/s. Block 2 is traveling

at 2 m/s in the same direction as Block 1. What is the final momentum and kinetic energy of Block 2?
A final momentum: 2 kg x m/s, final kinetic energy: 1 J
B. final momentum: 20 kg x m/s, final kinetic energy: 10 J
C final momentum: 90 kg x m/s, final kinetic energy: 20 J
D. final momentum: 22 kg x m/s, final kinetic energy: 101 J​
Physics
1 answer:
pentagon [3]3 years ago
4 0

Answer:

p = 20 kg•m/s

KE = 100 J

Explanation:

In an elastic collision of identical masses, the two masses will exchange momentums. Therefore Block 1 initially moving at 10 m/s will be moving at 2 m/s, and Block 2 will go from 2 m/s to 10 m/s

momentum = mv = 2(10) = 20 kg•m/s

KE = ½mv² = ½(2)10² = 100 J

Unfortunately, your answer selection does not have this answer as an option.

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Two balls with equal masses, m, and equal speed, v, engage in a head on elastic collision. what is the final velocity of each ba
Allushta [10]
The collision is elastic. This means that both momentum and kinetic energy are conserved after the collision.

- Let's start with conservation of momentum. The initial momentum of the total system is the sum of the momenta of the two balls, but we should put a negative sign in front of the velocity of the second ball, because it travels in the opposite direction of ball 1. So ball 1 has mass m and speed v, while ball 2 has mass m and speed -v:
p_i = p_1-p_2 = mv-mv =0
So, the final momentum must be zero as well:
p_f = 0
Calling v1 and v2 the velocities of the two balls after the collision, the final momentum can be written as
p_f = mv_1 + mv_2 = 0
From which
v_1 = -v_2

- So now let's apply conservation of kinetic energy. The kinetic energy of each ball is \frac{1}{2} mv^2. Therefore, the total kinetic energy before the collision is
K_i = \frac{1}{2} mv^2 +  \frac{1}{2} mv^2 = mv^2
the kinetic energy after the collision must be conserved, and therefore must be equal to this value:
K_f = K_i = mv^2 (1)
But the final kinetic energy, Kf, is also
K_f =  \frac{1}{2} mv_1^2 +  \frac{1}{2}mv_2^2
Substituting v_1 = -v_2 as we found in the conservation of momentum, this becomes
K_f = mv_2 ^2
we also said that Kf must be equal to the initial kinetic energy (1), therefore we can write 
mv_2^2 = mv^2

Therefore, the two final speeds of the balls are
v_2 = v
v_1 = -v_2 = -v

This means that after the collision, the two balls have same velocity v, but they go in the opposite direction with respect to their original direction.

8 0
3 years ago
How much heat is needed to boil 120 kg of water ?
nekit [7.7K]
Q = mc<span>∆t, where:
q = energy flow
m = mass, 120 000 g
c = specific heat capacity, 4.81 J/gC
</span><span>∆t = change in temperature, ~75 (100 - 25, which is room temperature)

Substituting in the values, we get:
q = 120000 x 4.81 x 75 = 43290000 Joules = 43.29 MJ

Hope I helped!! xx


</span>
7 0
3 years ago
An object of mass M oscillates on the end of a spring. To double the period, replace the object with one of mass: (a) 2M. (b)M/2
harkovskaia [24]

Answer:

(c) 4M

Explanation:

The system is a loaded spring. The period of a loaded spring is given by

T = 2\pi\sqrt{\dfrac{m}{k}}

<em>m</em> is the mass and <em>k</em> is the spring constant.

It follows that, since <em>k</em> is constant,

T\propto\sqrt{m}

\dfrac{T}{\sqrt{m}} = C where <em>C</em>  represents a constant.

\dfrac{T_1}{\sqrt{m_1}} = \dfrac{T_2}{\sqrt{m_2}}

m_2 = m_1\left(\dfrac{T_2}{T_1}\right)^2

When the period is doubled, T_2 = 2T_1.

m_2 = m_1\left(\dfrac{2T_1}{T_1}\right)^2 = 4m_1

Hence, the mass is replaced by 4M.

7 0
4 years ago
a positively charged body makes contact with a body. after a while, the charged body becomes neutralised. state 3 main condition
MrRa [10]

Answer:

A body will become positively charged when some electrons will come out from the body.Thus, positive charge is due to deficiency of electrons.

5 0
3 years ago
Computers are fast and do not make errors when doing calculations. Does this mean that computer models are always correct? Expla
Alika [10]
Computer models can have errors in them that need to be removed and fixed as they are discovered and they can also have glitches inside of them.
7 0
3 years ago
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