ANSWER:
a) 1.757 m/s
b) 119.91 N
STEP-BY-STEP EXPLANATION:
Given:
Mass 1 (m1) = 210 kg
Initial speed 1 (u1) = 3.5 m/s
Mass 2 (m2) = 221 kg
Initial speed (u2) = 1.8 m/s
We make a sketch of the situation:
a)
We make a momentum balance by taking into account the conservation of momentum:
[tex]\begin{gathered} m_1u_1-m_2u_2=m_1v_1+m_2v_2 \\ \\ v_2=\frac{m_1u_1-m_2u_2-m_1v_1}{m_2}\rightarrow(1) \end{gathered}[/tex]Now an energy balance taking into account the conservation of energy, as follows:
[tex]\begin{gathered} \frac{1}{2}m_1(u_1)^2+\frac{1}{2}m_2(u_2)^2=\frac{1}{2}m_1(v_1)^2+\frac{1}{2}m_2(v_2)^2 \\ \\ m_1\left(u_1\right)^2+m_2\left(u_2\right)^2=m_1\left(v_1\right)^2+m_2\lparen v_2)^2\rightarrow(2) \end{gathered}[/tex]Now, we substitute equation (1) in (2) and we get the following:
[tex]\begin{gathered} m_1\left(u_1\right)^2+m_2\left(u_2\right)^2=m_1\left(v_1\right)^2+m_2\left(\frac{m_1u_1-m_2u_2-m_1v_1}{m_2}\right)^2 \\ \\ m_1\left(u_1\right)^2+m_2\left(u_2\right)^2=m_1\left(v_1\right)^2+m_2\left(\frac{\left(m_1u_1\right)^2-2\left(m_1u_1\right)\left(m_2u_2\right)-2\left(m_1u_1\right)\left(m_1v_1\right)+\left(m_2u_2\right)^2+2\left(m_2u_2\right)\left(m_1v_1\right)+\left(m_1v_1\right)^2}{(m_2)^2}\right) \\ \\ m_1\left(u_1\right)^2+m_2\left(u_2\right)^2=m_1\left(v_1\right)^2+\left(\frac{\left(m_1u_1\right)^2-2\left(m_1u_1\right)\left(m_2u_2\right)-2\left(m_1u_1\right)\left(m_1v_1\right)+\left(m_2u_2\right)^2+2\left(m_2u_2\right)\left(m_1v_1\right)+\left(m_1v_1\right)^2}{m_2}\right) \\ \\ v_1=u_1\frac{m_1-m_2}{m_1+m_2}+u_2\frac{2m_2}{m_1+m_2} \\ \\ \text{ Now, we substitute each value, like so:} \\ \\ v_1=3.5\cdot\frac{210-221}{210+221}+1.8\cdot\frac{2\cdot221}{210+221} \\ \\ v_1=1.757\text{ m/s} \end{gathered}[/tex]b)
We use the following formula to determine the force:
[tex]\begin{gathered} v^2=u^2+2as \\ \\ \text{ Wee replacing:} \\ \\ (1.756)^2=0^2+2\cdot a\cdot2.7 \\ \\ a=0.003375\text{ m/s}^2 \\ \\ \text{ Therfore:} \\ \\ F=m\cdot a=210\cdot0.571=119.91\text{ N} \end{gathered}[/tex]
