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sashaice [31]
3 years ago
14

A 11,000-kg train car moving due east at 21.0 m/s collides with and couples to a 23,000-kg train car that is initially at rest.

What is the common velocty of the two-car train after the collisions?
Physics
1 answer:
Tomtit [17]3 years ago
6 0

Answer:

v = 6.79 m/s

Explanation:

It is given that,

Mass of a train car, m₁ = 11000 kg

Speed of train car, u₁ = 21 m/s

Mass of other train car, m₂ = 23000 kg

Initially, the other train car is at rest, u₂ = 0

It is a case based on inelastic collision as both car couples each other after the collision. The law of conservation of momentum satisied here. So,

m_1u_1+m_2u_2=(m_1+m_2)V

V is the common velocity after the collisions

V=\dfrac{m_1u_1}{(m_1+m_2)}\\\\V=\dfrac{11000\times 21}{(11000+23000)}\\\\V=6.79\ m/s

So, the two car train will move with a common velocity of 6.79 m/s.

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A cart loaded with bricks has a total mass of 9.13 kg and is pulled at constant speed by a rope. The rope is inclined at 24.7 ◦
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Work (W) is defined as the point product of force (F) by the distance (d)the body travels due to this force.  

W= F*d *cosα Formula (1)  

F : force (N)

d : displacement (m)

α : angle between force and displacement

Newton's second law:

∑F = m*a Formula (2)  

∑F : algebraic sum of the forces in Newton (N)

m : mass s (kg)

a : acceleration  (m/s²)

We define the x-axis in the direction parallel to the movement of the cart on the ramp and the y-axis in the direction perpendicular to it.

Forces acting on the cart

W: Weight of the cart : In vertical direction

FN : Normal force : perpendicular to the floor

f : Friction force: parallel to the floor

T : tension Force,  inclined at  θ=24.7° above the horizontal

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W= m*g

W= 9.13 kg* 9.8 m/s² = 89.47 N

x-y components o the  tension force (T)

Tx = Tcosθ = T*cos 24.7° (N)

Ty = Tsin θ = T*sin 24.7°  (N)

Calculated of the FN  

We apply the formula (2)  

∑Fy = m*ay ay = 0  

FN +Ty- W = 0  

FN = W-Ty  

FN =  89.47-T*sin 24.7°

Calculated of the friction force (f)

f = μk*FN

f =(0.597)*(  89.47-T*sin 24.7° )

f= 53.41-0.249T

Calculated of the tension force of the rope (f)

We apply the formula (2) :

∑Fx = m*ax  ,  ax= 0 ,because the speed of the cart  is constant

Tx - f = 0

T*cos 24.7°-( 53.41 - 0.249T )= 0

T*cos 24.7° + 0.249T = 53.41

(1.1575)T = 53.41

T= (53.41) / (1.1575)

T= 46.14 N

Work done on the cart by the rope

We apply the formula (1)

W=T*d *cosα

W= (46.14 N)*(15.1 m) *(cos24.7)

W = 632.97 (N*m) = 632.97 (J)

W = 0.63 KJ

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