Answer to Question #104373 in Classical Mechanics for Alize Magana

Question

You and your friend find yourselves in an unfortunate situation. A shopping-cart-related stunt has gone wrong and now your friend is stuck in a shopping cart and is on a crash course for your house with speed V= 1.78 m/s. You are on the roof of your house, which forms an angle θ=40.9 with the horizontal, and it's your job to stop your friend. You size up the situation and see that the mass of your friend and the cart combined is M=42.0 kg and luckily you have some camera equipment with you whose total mass is m=40.3 kg. Your plan is as follows. You're going to climb up the roof, place the equipment on the (frictionless) rooftoop, and let it slide off the roof, which is a height h=3.60 m above the ground. Your friend is going to catch the equipment and get slowed down by it. How far up the roof do you need to climb in order for your friend to come a complete stop?

Answer 1

As per the data given in the question,

Speed of the cart (v)=1.78 m/sec

angle of the roof $=40.9^\circ$

Mass of the man and the cart together (M)=42.0kg

Mass of the me and the camera equivalent together = 40.3 kg

distance covered by on the inclined plane is l,

height of the roof top(H)=3.6 m

Now, applying the conservation of momentum,

The energy of the cart and the man= $\dfrac{MV^2}{2}=\dfrac{42.0\times 1.78^2}{2}=66.5364J -------(i)$

The total energy of the me with camera with camera = $mg\sin\theta \times l$

$=40.3 \times9.8\sin40.9^\circ\times l -------(ii)$

from equation (i) and (ii)

$l=\dfrac{66.5364}{40.3\times9.8\times \sin(40.9^\circ)}$

$l=0.2573 m$

So the height raised by the cart and me $l \sin40.9^\circ=0.2573\times \sin(40.9^\circ) =0.1684 m$