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steposvetlana [31]

3 years ago

5

A 6.4 kg block with a speed of 5.4 m/s collides with a 12.8 kg block that has a speed of 3.6 m/s in the same direction. After th

e collision, the 12.8 kg block is observed to be traveling in the original direction with a speed of 4.5 m/s. (a) What is the velocity of the 6.4 kg block immediately after the collision

1 answer:

jenyasd209 [6]3 years ago

3 0

Answer:

The velocity of the 6.4 kg block immediately after the collision is 3.6 m/s in the original direction.

Explanation:

Given;

mass of first block, m₁ = 6.4 kg

initial speed of first block, u₁ = 5.4 m/s

mass of second block, m₂ = 12.8 kg

initial speed of second block, u₂ = 3.6 m/s

final speed of second block, v₂ = 4.5 m/s

To determine the final speed of 6.4 kg block immediately after the collision, we apply principle of conservation linear momentum;

Total momentum before collision = Total momentum after collision

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

where;

v₁ is the final speed of 6.4 kg block

v₂ is the final speed of 12.8 kg block

Substitute the given values of m₁, u₁, m₂, u₂ , v₂ and calculate v₁

6.4 x 5.4 + 12.8 x 3.6 = 6.4v₁ + 12.8 x 4.5

34.56 + 46.08 = 6.4v₁ + 57.6

80.64 = 6.4v₁ + 57.6

80.64 - 57.6 = 6.4v₁

23.04 = 6.4v₁

v₁ = 23.04 / 6.4

v₁ = 3.6 m/s

Therefore, the velocity of the 6.4 kg block immediately after the collision is 3.6 m/s in the original direction.

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Given that,

Mass of object = 9.90 kg

Time =5.40 s

Suppose the force is (2.00i + 9.00j + 5.30k) N, initial position is (2.70i - 2.90j + 5.50k) m and final position is (-4.10i + 3.30j + 5.40k) m.

We need to calculate the displacement

Using formula of displacement

$s=r_{2}-r_{1}$

Where, $r_{1}$ = initial position

$r_{2}$ = final position

Put the value into the formula

$s= (-4.10i + 3.30j + 5.40k)-(2.70i - 2.90j + 5.50k)$

$s= -6.80i+6.20j-0.1k$

(a). We need to calculate the work done on the object

Using formula of work done

$W=F\cdot s$

Put the value into the formula

$W=(2.00i + 9.00j + 5.30k)\cdot (-6.80i+6.20j-0.1k)$

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$W=41.67\ J$

(b). We need to calculate the average power due to the force during that interval

Using formula of power

$P=\dfrac{W}{t}$

Where, P = power

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Put the value into the formula

$P=\dfrac{41.67}{5.40}$

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(c). We need to calculate the angle between vectors

Using formula of angle

$\theta=\cos^{-1}(\dfrac{r_{1}r_{2}}{|r_{1}||r_{2}|})$

Put the value into the formula

$\theta=\cos^{-1}\dfrac{(-4.10i + 3.30j + 5.40k)\cdot(2.70i - 2.90j + 5.50k)}{7.54\times6.778})$

$\theta=79.7^{\circ}$

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g100num [7]

Answer:pounds

Explanation:

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Assuming a 8 kilogram bowling ball moving at 2 m/s bounces off a spring at the same speed that had before bouncing what is the a

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Assuming the spring is initially at rest, the total momentum of the system before the collision is given only by the momentum of the bowling ball:

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The ball bounces off at the same speed had before, but the new velocity has a negative sign (since the direction is opposite to the initial direction). So, the new momentum of the ball is:

$p_{fB}=m v_b =(8 kg)(-2 m/s)=-16 kg m/s$

The final momentum after the collision is the sum of the momenta of the ball and off the spring:

$p_f = p_{fB}+p_{fS}$

where $p_{fS}$ is the momentum of the spring. For the conservation of momentum,

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The change in momentum of bowling ball is given by the difference between its final momentum and initial momentum:

$\Delta p=p_{fb}-p_i=-16 kg m/s - 16 kg m/s=-32 kg m/s$

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Since we know $\Delta t=0.5 s$ , we can find the magnitude of the force:

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The negative sign simply means that the direction of the force is opposite to the initial direction of the ball.

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Thus, we can conclude that, when light is incident parallel to the principal axis and then strikes a lens, the light will refract through the focal point on the opposite side of the lens.

Learn more about refraction by a lens here:

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