Question

1. in a physics lab, a 2.0 kg cart (cart a) is moving rightward at 3.0 m/s when it collides head - on with a 1.0 kg cart (cart b) moving leftward at 2.0 m/s. after the collision, the two carts stick together and move as a single object. determine the post - collision velocity (include the direction!) of the combined carts. (1 point) select your answer

Explanation:

Step1: Define variables and choose formula

Let $m_1 = 2.0\ kg$, $v_1=3.0\ m/s$ (right - ward, so positive), $m_2 = 1.0\ kg$, $v_2=- 2.0\ m/s$ (left - ward, so negative). Use the conservation of momentum formula $m_1v_1 + m_2v_2=(m_1 + m_2)v_f$.

Step2: Substitute values into formula

$(2.0\ kg\times3.0\ m/s)+(1.0\ kg\times(-2.0\ m/s))=(2.0\ kg + 1.0\ kg)v_f$.
First, calculate the left - hand side:
$2.0\times3.0+1.0\times(-2.0)=6.0 - 2.0=4.0\ kg\cdot m/s$.
The right - hand side is $(2.0 + 1.0)v_f=3.0\ kg\cdot v_f$.

Step3: Solve for $v_f$

We have the equation $4.0\ kg\cdot m/s = 3.0\ kg\cdot v_f$.
So, $v_f=\frac{4.0\ kg\cdot m/s}{3.0\ kg}=\frac{4}{3}\ m/s\approx1.33\ m/s$. Since the value is positive, the direction is right - ward.

Answer:

The post - collision velocity is $\frac{4}{3}\ m/s$ or approximately $1.33\ m/s$ to the right.