A grocery shopper tosses a(n) 8.1 kg bag of rice into a stationary 18.2 kg grocery cart. The bag hits the cart with a horizontal speed of 6.9 m/s toward the front of the cart. What is the final speed of the cart and bag?

2.125 m/s

Explanation:

From the law of conservation of momentum,

Total momentum before collision = Total momentum after collision.

Note: Assuming a inelastic collision occur between the grocery cart and the bag.

mv+m'v' = V(m+m') Equation 1

Where m = mass of the cart, m' = mass of the bag, v = initial velocity of the grocery cart, v' = final velocity of the bag, V = Final velocity of the bag and the cart.

Making V the subject of the equation,

V = (mv+m'v')/(m+m') Equation 2

Given: m = 18.2 kg, m' = 8.1 kg, v' = 6.9 m/s, v = 0 m/s (stationary)

Substitute into equation 2

V = [(18.2×0)+(8.1×6.9)]/(8.1+18.2)

V = 55.89/26.3

V = 2.125 m/s.

Hence the final speed of the cart and the bag = 2.125 m/s

0.863333

86%

hope this

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