Question

Tarzan is in the path of a pack of stampeding elephants when Jane swings in to the rescue on a rope vine, hauling him off to safety. The length of the vine is 27 m, and Jane starts her swing with the rope horizontal. If Jane's mass is 49 kg, and Tarzan's mass is 86 kg, to what height above the ground will the pair swing after she rescues him

Answer 1

Answer:

h = 3.56 m

Explanation:

• Assuming no friction on the rope or air resistance, we can apply the conservation of mechanical energy principle to the first part of the trajectory, from when Jane starts her swing till she catches Tarzan at the ground level.

⇒      $\Delta K + \Delta U = 0 (1)$

• Rearranging terms, we get:

        $U_{ij} + K_{ij} = U_{fj} + K_{f} (2)$

• Now, as Jane starts from rest, Kij =0.
• if we choose the ground level as our zero reference level, it will be also  Ufj = 0.
• Replacing in (2) by the expressions of Uij and Kfj, we have:

        $m_{j} * g* h_{ij} = \frac{1}{2} * m_{j} * v_{fj} ^{2} (3)$

• Replacing in (3) by the givens, and rearranging terms, we can solve for vfj, as follows:

       $v_{fj} =\sqrt{2*9.8 m/s2*27 m} = 23 m/s (4)$

• Now, as once jane catches Tarzan, both continue swinging together, we can take the catching moment as a completely inelastic collision.
• Assuming no external forces act during the collision, total momentum must be conserved.

⇒     $p_{o} = p_{f} (5)$

• Assuming that Tarzan is at rest when Jane catches him, the initial momentum will be simply as follows:

        $p_{o} = m_{j} * v_{j} = 49 kg * 23 m/s = 1127 kg*m/s (6)$

• The final momentum, will be just the product of the combined mass of Jane and Tarzan times the common speed for them after the collision:

        $p_{f} = (m_{j} + m_{t} ) * v_{jt} = 135 kg* v_{jt} (7)$

• As (6) and (7) are equal each other, we can solve for vjt, as follows:

       $v_{jt} = \frac{1127 kg*m/s}{135 kg} = 8.35 m/s (8)$

• Finally we can apply the same energy conservation principle to the last part of the trajectory, as follows:

       $U_{ijt} + K_{ijt} = U_{fjt} + K_{fjt} (9)$

• We know that Uijt = 0 and also that Kfjt = 0, due to both starts from the ground level and reach to the highest point before starting to fall down, so at this point, the kinetic energy will be zero.
• Replacing by the givens and the result from (8), we can solve for the h as follows:
• $h_{f} =\frac{v_{ijt} ^{2} }{2*g} = \frac{69.7m2/s2}{2*9.8m/s2} = 3.56 m$