Respuesta :
Consider the law of conservation of momentum in the given situation.
(a) The momentum po of the system before the child jumps to the sled is equal to the total momentum after pf. Then, you have:
po = pf
po = m1*v1
pf = (m1 + m2)*v
where,
m1: mass of the child = 50kg
m2: mass of the sled = 10kg
v1: speed of the child = 0.6m/s
v: speed of the child and the sled togueter = ?
Solve the equation above for v and replace the previous values of the parameters:
[tex]\begin{gathered} m_1v_1=(m_1+m_2)v \\ v=\frac{m_1v_1}{m_1+m_2} \\ v=\frac{(50kg)(0.6\frac{m}{s})}{50\operatorname{kg}+10\operatorname{kg}} \\ v=0.5\frac{m}{s} \end{gathered}[/tex]The speed of the system child-sled is 0.5m/s
(b) The kinetic energy is given by:
[tex]\begin{gathered} K=\frac{1}{2}(m_1+m_2)v^2 \\ K=\frac{1}{2}(50kg+10kg)(0.5\frac{m}{s})^2 \\ K=7.5J \end{gathered}[/tex]The kinetic energy of the child-system is 7.5J
(c) Again, the total momentum must conserve before and after the child jumps backward off the sled. If after the jump the child is at rest on the snow, you have:
po = pf
po = (m1 + v1)*v
pf = m2*v2
where,
v2: speed of the sled after the child jumps backward off the sled = ?
equal the expressions for po and pf, solve for v2 and replace the values of the rest of the paramters:
[tex]\begin{gathered} (m_1+m_2)v=m_2v_2 \\ v_2=\frac{(m_1+m_2)v}{m_2} \\ v_2=\frac{(50kg+10kg)(0.5\frac{m}{s})}{10\operatorname{kg}} \\ v_2=3\frac{m}{s} \end{gathered}[/tex]After the child lands the speed of the sled is 3m/s
(d) The kinetic energy is given by:
[tex]\begin{gathered} K=\frac{1}{2}m_2v^2_2 \\ K=\frac{1}{2}(10kg)(3\frac{m}{s})^2 \\ K=45J \end{gathered}[/tex]The kinetic energy of the sled is 45J
