To find the final speed of the cart and the bag after the collision, we can use the principle of conservation of momentum. The momentum before the collision must equal the momentum after the collision.
Let:
- \( m_1 = 9.0 \) kg (mass of the bag of rice)
- \( v_1 = 5.8 \) m/s (velocity of the bag of rice before the collision)
- \( m_2 = 19.8 \) kg (mass of the grocery cart)
- \( v_2 = 0 \) m/s (velocity of the cart before the collision)
- \( v_f \) = final velocity of the cart and bag after the collision
The total initial momentum (before the collision) is given by: \[ p_{\text{initial}} = m_1 v_1 + m_2 v_2 \] Substituting in the values: \[ p_{\text{initial}} = (9.0 , \text{kg})(5.8 , \text{m/s}) + (19.8 , \text{kg})(0 , \text{m/s}) = 52.2 , \text{kg m/s} \]
The total final momentum (after the collision) is given by: \[ p_{\text{final}} = (m_1 + m_2)v_f \] Setting the initial momentum equal to the final momentum: \[ 52.2 , \text{kg m/s} = (9.0 , \text{kg} + 19.8 , \text{kg}) v_f \] \[ 52.2 , \text{kg m/s} = (28.8 , \text{kg}) v_f \]
Now, solving for \( v_f \): \[ v_f = \frac{52.2 , \text{kg m/s}}{28.8 , \text{kg}} \approx 1.8125 , \text{m/s} \]
Therefore, rounding to three significant figures, the final speed of the cart and bag together is: \[ \boxed{1.81 , \text{m/s}} \]
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