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Lubov Fominskaja [6]
3 years ago
14

wo skaters collide and embrace in an inelastic collision. Alex's mass is 90 kg and his initial velocity is 1.5 m/s i . Barbara's

mass is 57 kg and her initial velocity is 2.0 m/s j. After the collision, the two skaters move together at a common velocity. Use conservation of momentum to find their final speed. (Hint, find combined velocity vector first.)
Physics
2 answers:
Vitek1552 [10]3 years ago
7 0

Answer:

1.21 m/s

Explanation:

mass of Alex, mA = 90 kg

initial velocity of Alex, uA = 1.5 i m/s

mass of Barbara, mB = 57 kg

initial velocity of Barbara, uB = 2 j m/s

Let v is the velocity of combined mass after collision.

Use conservation of momentum

90\times 1.5\widehat{i}+57\times 2\widehat{j}=(90+57)\overrightarrow{v}

\overrightarrow{v} = 0.92\widehat{i}+0.78\widehat{j}

v=\sqrt{0.92^{2}+0.78^{2}}

v = 1.21 m/s

Thus, the velocity of combined mass is 1.21 m/s.

Natalija [7]3 years ago
3 0

Answer:

The two skaters will move with a common speed of 1.19 m/s.

Explanation:

Given that,

Mass of Alex, m_1=90\ kg

Initial velocity of Alex, u_1=1.5i\ m/s

Mass of Barbara, m_2=57\ kg

Initial velocity of Barbara, u_2=2j\ m/s

After the collision, the two skaters move together at a common velocity. Let V is the common velocity. Using the conservation of momentum as :

m_1u_1+m_2u_2=(m_1+m_2)V\\\\90\times 1.5i+57\times 2j=(90+57)V\\\\V=\dfrac{135i+114j}{147}\\\\V=\dfrac{135i}{147}+\dfrac{114j}{147}\\\\V=(0.91i+0.77j)\ m/s

Magnitude of final velocity:

|V|=\sqrt{0.91^2+0.77^2} \\\\|V|=1.19\ m/s

So, the two skaters will move with a common speed of 1.19 m/s.

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A steel beam of mass 1975 kg and length 3 m is attached to the wall with a pin that can rotate freely on its right side. A cable
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Answer:

a) 29062.125 N·m

b) 0 N·m

c) Torque, due \ to \ tension =L\cdot Tsin\theta = \frac{M\cdot L\cdot g}{2}

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Explanation:

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Beam mass = 1975 kg

Beam length = 3 m

Cable angle = 60° above horizontal

a) We have the formula for torque given as follows;

Torque about the pin = Force × Perpendicular distance of force from pin

Where the force = Force due to gravity or weight, we have

Weight = Mass × Acceleration due to gravity = 1975 × 9.81 = 19374.75 N

Point of action of force = Midpoint for a uniform beam = length/2

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Torque due to gravity = 19374.75 N × 1.5 m = 29062.125 N·m

b) Torque about the pinned end due to the contact forces between the pin and the beam is given by the following relation;

Since the distance from pin to the contact forces between the pin and the beam is 0, the torque which is force multiplied by perpendicular distance is also 0 N·m

c) To find the expression for the tension force, T we find the sum of the moment forces about the pin as follows

Sum of moments about p is given as follows

∑M = 0 gives;

T·sin(θ) × L= M×L/2×g

Therefore torque due to tension is given by the following expression

Torque, due \ to \ tension =L\cdot Tsin\theta = \frac{M\cdot L\cdot g}{2}

d) Plugging in the values in the torque due to tension equation, we have;

3\times Tsin60 = \frac{1975\times 3\times 9.81}{2} = 29062.125

Therefore, we make the tension force, T the subject of the formula hence

T= \frac{29062.125}{3 \times sin(60)} = 11186.02 N

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