A grocery shopper tosses a 9.0 kg bag of rice into a stationary 18.0 kg grocery cart. The bag hits the cart with a horizontal speed of 5.5 m/s toward the front of the cart. What is the final speed of the cart and bag

The speed with which the cart and bag move after the collision is equal to 1.83 m/s.

Given the following data:

To calculate the final speed with which the cart and bag move after the collision:

Applying the law of conservation of momentum, the collision between the cart and bag is given by the formula:

Where:

Substituting the given parameters into the formula, we have;

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1.83m/s

Explanation:

According to the law of conservation of momentum which States that "the sum of momentum of bodies before collision is equal to the sum of momentum of the bodies after collision". The objects moves with a common velocity (v) after collision.

Since momentum = mass × velocity

Let m1 and m2 be the masses of the bag and the cart respectively and;

u1 and u2 be the velocities of the bag and the cart respectively.

v is the common velocity

momentum of the bag of rice before collision= m1u1

= 9×5.5

= 49.5kgm/s²

momentum of the cart before collision= m2u2

Since the grocery cart is stationary, its velocity will be 0m/s

Its momentum = 18(0) = 0kgm/s

Final momentum of the bodies after collision = (m1+m2)v

= (9+18)v

= 27v kgm/s

According to the law of conservation of momentum;

49.5+0 = 27v

v = 49.5/27

v = 1.83m/s

Therefore the final speed of the cart and bag is 1.83m/s

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