A gymnast of mass 65 kg bounces vertically on a trampoline so thatshe approaches and leaves the trampoline with a speed of 10 m/s andcontact time of 0.9 s. Calculate (1) her change of momentum (ii) theaverage resultant force exerted on her while in contact with thetrampoline​

Answer:Change in momentum: [tex]1.3 \times 10^{3}\;\rm kg \cdot m \cdot s^{-1}[/tex]Average resultant force: approximately [tex]1.4 \times 10^{3}\; \rm N[/tex]. Explanation:Consider an object of mass [tex]m[/tex] travelling at velocity [tex]v[/tex]. The momentum of this object would be [tex]p = m \cdot v[/tex]. Momentum is a vector. The direction of an object's momentum is the same as the direction of its velocity.Initial velocity of this gymnast: [tex]v_0 = -10\; \rm m \cdot s^{-1}[/tex]. This value is negative because the gymnast was initially travelling downwards.Hence, the initial momentum of this gymnast would be:[tex]\begin{aligned}p_0 &= m \cdot v_0 \\ &= 65\; \rm kg \times (-10\; \rm m \cdot s^{-1})\\ &= -650\; \rm kg \cdot m \cdot s^{-1}\end{aligned}[/tex].Velocity of this gymnast after leaving the trampoline: [tex]v_1 = 10\; \rm m\cdot s^{-1}[/tex].Hence, the momentum of this gymnast when leaving the trampoline would be: [tex]\begin{aligned}p_1 &= m \cdot v_1 \\ &= 65\; \rm kg \times 10\; \rm m \cdot s^{-1}\\ &= 650\; \rm kg \cdot m \cdot s^{-1}\end{aligned}[/tex].Change in momentum:[tex]\begin{aligned}&p_1 - p_0 \\ &= 650\; \rm kg \cdot m \cdot s^{-1} - (-650\; \rm kg \cdot m \cdot s^{-1}) \\ &= 1300\; \rm kg \cdot m\cdot s^{-1}\end{aligned}[/tex].By Newton's Second Law of motion, if the momentum of an object changed by [tex]\Delta p[/tex] over time [tex]t[/tex], the average net force on that object would be:[tex]\begin{aligned}F &= \frac{\Delta p}{t}\end{aligned}[/tex].For this gymnast:[tex]\begin{aligned}F&= \frac{\Delta p}{t} \\ &= \frac{1300\; \rm kg \cdot m \cdot s^{-1}}{0.9\; \rm s} \\ &\approx 1.4 \times 10^{3}\; \rm kg \cdot m \cdot s^{-2} \\ &\approx 1.4 \times 10^{3}\; \rm N\end{aligned}[/tex].