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lapo4ka [179]
3 years ago
5

Two blocks are on a horizontal frictionless surface. Block a is moving with an initial velocity

Physics
1 answer:
AfilCa [17]3 years ago
5 0

The final velocity of the block A will be 2.5 m/sec. The principal of the momentum conversation is used in the given problem.

<h3>What is the law of conservation of momentum?</h3>

According to the law of conservation of momentum, the momentum of the body before the collision is always equal to the momentum of the body after the collision.

In a given concern, mass m₁ is M, mass m₂ is 3M. Initial speed for the mass m₁ and  m₂ will be u₁=5 and u₂=0 m/s respectively,

According to the law of conservation of momentum

Momentum before collision =Momentum after collision

m₁u₁+m₂u₂=(m₁+m₂)v

M×5+3M×0=[M+3M]v

The final velocity is found as;

V=51.25 m/s

The velocity of block A is found as;

\rm V_f=\frac{(m_1-m_2)u_1}{m_1+m_2} +\frac{2m_2u_2}{m_1+m_2} \\\\  V_f=\frac{(M-3M)5+2\times3M\times0}{m_1+m_2} \\\\ V_f=2.5 m/s

Hence, the final velocity of the block A will be 2.5 m/sec.

To learn more about the law of conservation of momentum, refer;

brainly.com/question/1113396

#SPJ4

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Assume that the upward direction is positive and the downward direction is negative. What is the ball's velocity (in m/s) when i
LenaWriter [7]

The given question is incomplete. The complete question is as follows.

You throw a ball vertically upward, and as it leaves your hand, its speed is 26.0 m/s.

(a) How high (in m) does it rise above the level where it leaves your hand?

(b) How long (in s) does it take to reach its highest point?

(c) How long (in s) does the ball take to return to the level where it left your hand after it reaches its highest point?

(d) Assume that the upward direction is positive and the downward direction is negative. What is the ball's velocity (in m/s) when it returns to the level where it left your hand? (Indicate the direction with the sign of your answer.)

Explanation:

(a) For maximum height, the formula will be as follows.

           v^{2} = u^{2} + 2as

                 a = v^{2} - 2gh

or,                 h = \frac{v^{2}}{2g}

                        = \frac{(26)^{2}}{2 \times 9.8}

                        = \frac{676}{19.6}

                        = 34.5 m/s

Hence, it rises 34.5 m/s above the level where it leaves your hand.

(b) Time to reach maximum height is as follows.

            v = u + at

or,           v - gt = 0

                 t = \frac{v}{g}

                   = \frac{26}{10}

                   = 2.6 sec

Therefore, it will take 2.6 sec to reach its highest point.

(c)  Time taken by the ball to ascent is equal to the time it has taken to descent.

Therefore, time taken by the ball to return to the level where it left your hand after it reaches its highest point? is also 2.6 sec.

(d)  Speed of the ball will be 26 m/s in the downward direction. Hence, the velocity will be -26 m/s.

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3 years ago
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A psychologist will need to have good linguistic intelligence in other to be successful.

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2 years ago
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SOLUTION is given in attachment below.

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3 years ago
If the range of a projectile's trajectory is six times larger than the height of the trajectory, then what was the angle of laun
zvonat [6]

Answer:

H = 1/2 g t^2    where t is time to fall a height H

H = 1/8 g T^2   where T is total time in air  (2 t  = T)

R = V T cos θ       horizontal range

3/4 g T^2 = V T cos θ       6 H = R    given in problem

cos θ = 3 g T / (4 V)           (I)

Now t = V sin θ / g     time for projectile to fall from max height

T = 2 V sin θ / g

T / V = 2 sin θ / g

cos θ = 3 g / 4 (T / V)     from (I)

cos θ = 3 g / 4 * 2 sin V / g = 6 / 4 sin θ

tan θ = 2/3      

θ = 33.7 deg

As a check- let V = 100 m/s

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R = 83.2 * 11.3 = 932

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