A grocery shopper tosses a(n) 9.4 kg bag of
rice into a stationary 19.9 kg grocery cart.
The bag hits the cart with a horizontal speed
of 7.0 m/s toward the front of the cart.
What is the final speed of the cart and bag?
Answer in units of m/
To find the final speed of the cart and bag, we can use the principle of conservation of momentum. The total momentum before the collision is equal to the total momentum after the collision.
The momentum of an object is given by the product of its mass and velocity:
Momentum = mass × velocity
Before the collision, the bag of rice has a mass of 9.4 kg and a velocity of 7.0 m/s. The cart has a mass of 19.9 kg and is stationary (0 m/s). Therefore, the total momentum before the collision is:
Total momentum before = (mass of bag × velocity of bag) + (mass of cart × velocity of cart)
= (9.4 kg × 7.0 m/s) + (19.9 kg × 0 m/s)
= 65.8 kg·m/s + 0 kg·m/s
= 65.8 kg·m/s
After the collision, the bag and cart move together with a common final velocity. Let's call this final velocity "v". The total mass of the bag and cart combined is the sum of their individual masses:
Total mass after = mass of bag + mass of cart
= 9.4 kg + 19.9 kg
= 29.3 kg
Using the principle of conservation of momentum, the total momentum after the collision is:
Total momentum after = (total mass after) × (final velocity)
= 29.3 kg × v
Since the total momentum before the collision is equal to the total momentum after the collision, we can set up the following equation:
Total momentum before = Total momentum after
65.8 kg·m/s = 29.3 kg × v
Now we can solve for the final velocity "v":
v = (65.8 kg·m/s) / (29.3 kg)
v ≈ 2.25 m/s
Therefore, the final speed of the cart and bag is approximately 2.25 m/s.