Question

two carts having the same mass of 10kg collide. cart 1 was traveling 1m/s, cart 2 was at rest. if the collision in elastic, how fast was each cart moving after the collision? cart 1 = 0.5 m/s cart 2 = 0.5 m/s; cart 1 = 1 m/s cart 2 = 0 m/s; cart 1 = 0 m/s cart 2 = 1 m/s; cart 1 = 0 m/s cart 2 = 10 m/s

Explanation:

Step1: Apply conservation of momentum

The formula for conservation of momentum is $m_1u_1 + m_2u_2=m_1v_1 + m_2v_2$. Given $m_1 = m_2=10$ kg, $u_1 = 1$ m/s, $u_2 = 0$ m/s. So, $10\times1+10\times0 = 10v_1+10v_2$, which simplifies to $10 = 10v_1 + 10v_2$ or $v_1 + v_2=1$.

Step2: Apply conservation of kinetic - energy for elastic collision

The formula for conservation of kinetic - energy in an elastic collision is $\frac{1}{2}m_1u_1^{2}+\frac{1}{2}m_2u_2^{2}=\frac{1}{2}m_1v_1^{2}+\frac{1}{2}m_2v_2^{2}$. Substituting the values: $\frac{1}{2}\times10\times1^{2}+\frac{1}{2}\times10\times0^{2}=\frac{1}{2}\times10\times v_1^{2}+\frac{1}{2}\times10\times v_2^{2}$, which simplifies to $10 = 10v_1^{2}+10v_2^{2}$ or $v_1^{2}+v_2^{2}=1$.

Step3: Solve the system of equations

From $v_1 + v_2 = 1$, we have $v_2=1 - v_1$. Substitute $v_2 = 1 - v_1$ into $v_1^{2}+v_2^{2}=1$: $v_1^{2}+(1 - v_1)^{2}=1$. Expand: $v_1^{2}+1 - 2v_1+v_1^{2}=1$. Combine like - terms: $2v_1^{2}-2v_1=0$. Factor out $2v_1$: $2v_1(v_1 - 1)=0$. So, $v_1 = 0$ m/s or $v_1 = 1$ m/s. If $v_1 = 0$ m/s, then from $v_1 + v_2=1$, $v_2 = 1$ m/s.

Answer:

C. Cart 1=0 m/s, Cart 2=1 m/s