<span>The 2nd truck was overloaded with a load of 16833 kg instead of the permissible load of 8000 kg.
The key here is the conservation of momentum.
For the first truck, the momentum is
0(5100 + 4300)
The second truck has a starting momentum of
60(5100 + x)
And finally, after the collision, the momentum of the whole system is
42(5100 + 4300 + 5100 + x)
So let's set the equations for before and after the collision equal to each other.
0(5100 + 4300) + 60(5100 + x) = 42(5100 + 4300 + 5100 + x)
And solve for x, first by adding the constant terms
0(5100 + 4300) + 60(5100 + x) = 42(14500 + x)
Getting rid of the zero term
60(5100 + x) = 42(14500 + x)
Distribute the 60 and the 42.
60*5100 + 60x = 42*14500 + 42x
306000 + 60x = 609000 + 42x
Subtract 42x from both sides
306000 + 18x = 609000
Subtract 306000 from both sides
18x = 303000
And divide both sides by 18
x = 16833.33
So we have the 2nd truck with a load of 16833.33 kg, which is well over it's maximum permissible load of 8000 kg. Let's verify the results by plugging that mass into the before and after collision momentums.
60(5100 + 16833.33) = 60(21933.33) = 1316000
42(5100 + 4300 + 5100 + 16833.33) = 42(31333.33) = 1316000
They match. The 2nd truck was definitely over loaded.</span>
For speed you can differentiate the equation, for acceleration you can again differentiate the equation .
at t=0 the particle is slowing down , when you get equation for velocity put t=0 then only -1 is left
Answer:
<em>B. 68.6m</em>
Explanation:
<u>Free Fall Motion
</u>
When a body is left to move in the air with no friction, the motion is ruled only by the force of gravity. The vertical distance a body travels in the air after a time t is
.

We know the egg takes 3.74 seconds to reach the ground. The height it was launched from is


The closest correct option is
B. 68.6m
Given that force is applied at an angle of 30 degree below the horizontal
So let say force applied if F
now its two components are given as


Now the normal force on the block is given as



now the friction force on the cart is given as



now if cart moves with constant speed then net force on cart must be zero
so now we have




so the force must be 199.2 N
<em>A</em> - <em>B</em> = (10<em>i</em> - 2<em>j</em> - 4<em>k</em>) - (<em>i</em> + 7<em>j</em> - <em>k</em>)
<em>A</em> - <em>B</em> = 9<em>i</em> - 9<em>j</em> - 3<em>k</em>
|<em>A</em> - <em>B</em>| = √(9² + (-9)² + (-3)²) = √189 = 3√19