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Ray Of Light [21]
3 years ago
8

The 5-kg block A has an initial speed of 5 m/s as it slides down the smooth ramp, after which it collides with the stationary bl

ock B of mass 8 kg. If the two blocks couple together after collision, determine their common velocity immediately after collision.
Physics
1 answer:
Akimi4 [234]3 years ago
8 0

Answer:

The coupled velocity of both the blocks is 1.92 m/s.

Explanation:

Given that,

Mass of block A, m_1=5\ kg

Initial speed of block A, u_1=5\ m/s

Mass of block B, m_2=8\ kg

Initial speed of block B, u_2=0

It is mentioned that if the two blocks couple together after collision. We need to find the common velocity immediately after collision. We know that due to coupling, it becomes the case of inelastic collision. Using the conservation of linear momentum. Let V is the coupled velocity of both the blocks. So,

m_1u_1+m_2u_2=(m_1+m_2)V\\\\V=\dfrac{m_1u_1+m_2u_2}{(m_1+m_2)}\\\\V=\dfrac{5\times 5+0}{(5+8)}\\\\V=1.92\ m/s

So, the coupled velocity of both the blocks is 1.92 m/s. Hence, this is the required solution.

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A 8.00 kg block of ice, released from rest at the top of a 1.50-m-long frictionless ramp, slides downhill, reaching a speed of 2
IRINA_888 [86]

We have that the speed of the ice at the bottom  is mathematically given as

v=1.8812m/s

<h3>The speed of the ice at the bottom </h3>

Question Parameters:

A 8.00 kg block of ice, released from rest

at the top of a 1.50-m-long frictionless ramp,

slides downhill, reaching a speed of 2.70 m/s

constant friction force of 10.0 N parallel to the surface of the ramp

Generally the equation for the Potential Energy is  is mathematically given as

P.E=K.E+W_{fric}

Therefore

mgh=0.5mv^2+W_{fric}

Where

W_{fric}=F.S

W_{fric}=10*1.50

W_{fric}=15

And

mgh=0.5mv^2+

h=0.3719

Hence

0.5(8.00)(v^2)=(8.00)(9.8)(0.3719)-15

v^2=3.53924

v=1.8812m/s

For more information on Velocity visit

brainly.com/question/17617636

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a bicycle rider pedals towards a hill at a speed of 7.5w/s. the combined mass of the bicycle and the rider is 75.0kg the cycle a
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I will assume 7.5 m/s

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KE = PE

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1/2 v^2 = gh         looking for  h     so divide both sides by

1/2 v^2 / g   = h      sub in the values   using g = 10 m/s^2

1/2 ( 7.5 )^2 / 10  = 2.8125 m

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