A proposed space elevator would consist of a cable stretching from the earth's surface to a satellite, orbiting far in space, th

Response:

The man's speed is 0.144 m/s

Explanation:

This exemplifies conservation of momentum.

The momentum of the ball prior to being caught must equal the momentum of the man-ball system after catching the ball.

Mass of the ball = 0.65 kg

Mass of the man = 54 kg

Speed of the ball = 12.1 m/s

The momentum of the ball before impact can be calculated as mass multiplied by velocity

= 0.65 x 12.1 = 7.865 kg-m/s

After catching the ball, the momentum of the combined system is

(0.65 + 54)Vf = 54.65Vf

Where Vf denotes their final shared velocity.

Setting the initial momentum equal to the final momentum,

7.865 = 54.65Vf

Vf = 7.865/54.65 = 0.144 m/s

The complete question is;

Block 1 sits on the floor with block 2 resting atop it. Block 3, which is stationary on a frictionless table, is attached to block 2 via a string that passes over a pulley depicted in the illustration below. Both the string and pulley have negligible mass.

Once block 1 is taken away without impacting block 2.

Derive an equation for the acceleration of block 3 considering arbitrary values for m3 and m2. Express your answer in terms of m3, m2, and relevant physical constants as needed.

Answer:

a = (m2)g/(m3 + m2)

Explanation:

Examining the attached illustration, by analyzing the free body diagram for block 3 and utilizing Newton's first law of motion, we reach the following formula;

T = (m3)a - - - (eq 1)

where;

T is the tension in the string

a is acceleration

m3 is the mass of block 3

Simultaneously, doing the same for Block 2, the free body diagram yields the equation; (m2)g - T = (m2)a

Rearranging for T results in;

T = (m2)g - (m2)a - - - (eq 2)

where;

g represents acceleration due to gravity

T is the tension in the string

a is acceleration

m2 is the mass of block 2

To deduce the acceleration, we will substitute (m3)a in place of T in eq 2.

Thus;

(m3)a = (m2)g - (m2)a

(m3)a + (m2)a = (m2)g

a(m3 + m2) = (m2)g

a = (m2)g/(m3 + m2)

Answer:

x₂=2×1

Explanation:

According to the work-energy theorem, we can assume that the gravitational potential energy at the lowest point of compression is zero since the kinetic energy change is 0;

mgx-(kx)²/2 =0 where m refers to the object's mass, g indicates the acceleration due to gravity, k denotes spring constant, and x represents the spring's compression.

mgx=(kx)²/2

x=2mg/k----------------compression when the object is at rest

However, ΔK.E =-1/2mv²⇒kx²=mv² -----------where v symbolizes the object's velocity and K.E signifies kinetic energy

Thus, if kx²=mv² then

v=x *√(k/m) ----------------where v=0