Atrolley of mass 200kg is travelling north at 2m\s. it collides head on with a second trolley of mass 500kg which is travelling south with a velocity of 5 m\s and they remain coupled together.
a)calculate the velocity of the two trolleys after collision.
b)what is the direction of motion of the trolley after collison?

The trolley has a velocity of 4.14 m/s west of north.

Recall that from the principle of conservation of linear momentum;

Momentum before collision = Momentum after collision

Let the mass of the first trolley be M1

Let the mass of the second trolley be M2

Let the initial velocity of the first trolley be U1

Let the final velocity of the first trolley be V1

Let the initial velocity of the second trolley be S1

Let the final velocity of the second trolley be S2

Given that;

M1U1 + M2S1 = M1V1 + M2S2

Since collision is inelastic

M1U1 + M2S1 = (M1 + M2)V

V = M1U1 + M2S1/(M1 + M2)

V =(200kg ×  2m\s) + (500kg × 5 m\s)/(200kg + 500kg)

V = 400 + 2500/700

V = 4.14 m/s

The direction of the trolleys after collision is west of north.

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what is the question?

weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards