Chapter 9: Q34P (page 249)

Basilisk lizards can run across the top of a water surface (Figure 9-52). With each step, a lizard first slaps its foot against the water and then pushes it down into the water rapidly enough to form an air cavity around the top of the foot. To avoid having to pull the foot back up against water drag in order to complete the step, the lizard withdraws the foot before water can flow into the air cavity. If the lizard is not to sink, the average upward impulse on the lizard during this full action of slap, downward push, and withdrawal must match the downward impulse due to the gravitational force. Suppose the mass of a basilisk lizard is 9.00 g, the mass of each foot is 3.00 g , the speed of a foot as it slaps the water is 1.50 m/s , and the time for a single step is 0.600 s .(a) What is the magnitude of the impulse on the lizard during the slap? (Assume this impulse is directly upward.) (b) During the 0.600 sduration of a step, what is the downward impulse on the lizard due to the gravitational force? (c) Which action, the slap or the push, provides the primary support for the lizard, or are they approximately equal in their support?

Short Answer

Expert verified

The magnitude of impulse on the lizard during the slap, $J is 4.50 \times 10^{- 3} N . s$ .
The downward impulse on lizard due to gravitational force, $J is 0.529 N . s$ .
The slap provides primary support for the lizard.

Step by step solution

Understanding the given information

The mass of the lizard, $m_{1} = 90.0 gm = 90.0 \times 10^{- 3} kg$ .
Mass of the lizard’s foot $m_{2} = 3.00 gm = 3.00 \times 10^{3} kg$ .
The speed of the slap, $v = 1.50 m / s$ .
The duration of slap, $∆ t = 0.600 s$ .

Concept and formula used in the given question

While walking on the surface, the lizard makes use of the impulse during the collision of feet with the water surface. Hence, you use the impulse-linear momentum theorem to determine the impulse due to slap and gravitational force which is given as.

$\vec{J} = ∆ \vec{p} = m ∆ \vec{v} = F_{avg} ∆ t$

Calculation for the magnitude of the impulse on the lizard during the slap

(a)

The magnitude of impulse during the slap is

$J = m ∆ v = 3.00 \times 10^{- 3} \times 1.50 = 4.50 \times 10^{- 3} N . s$

Calculation for the downward impulse on the lizard due to the gravitational force

(b)

The magnitude of the impulse due to gravitational force is calculated as

$J = F_{a v g} ∆ t = F_{a v g} ∆ t = m g ∆ t = 9.00 \times 10^{- 3} \times 9.8 \times 0.600 = 0.529 N . s$

Calculation for the action, the slap or the push, which provides the primary support for the lizard, or are they approximately equal in their support

(c)

The push provides the primary support to the lizard. If we compare the impulse due to slap with the impulse due to gravity, gravitational impulse is more in magnitude. Hence, it is the push that tries to balance the impulse due to gravity.