Two children stand on a platform at the top of a curving slide next to a backyard swimming pool. At the same moment the smaller child hops off to jump straight down into the pool, the bigger child releases herself at the top of the frictionless slide.(i) Upon reaching the water, how does the kinetic energy of the smaller child compare with that of the larger child?a. It is greater. It is less.
b. The two energies are equal. (ii) Upon reaching the water, how does the speed of the smaller child compare with that of the larger child is?
a. It is greater. It is less.
b. The two speeds are equal. (iii) During their motions from the platform to the water, how does the average acceleration of the smaller child compare with that of the larger child?
a. It is greater.
b. It is less.
c. The two accelerations are equal.d. The First and third are wrong

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aristocles aristocles · Answer 1 · 2019-11-07T02:35:24+01:00

1.

Answer:

a) It is less

Explanation:

By energy conservation we can say that initial potential energy of both child must be equal to the final kinetic energy of the two child.

Since initially they are at same height so we will say that initial potential energy will be given as

[tex]mgH [/tex] and MgH

so the child with greater mass has more energy and hence smaller child will reach with smaller kinetic energy

2.

Answer:

b. The two speeds are equal.

Explanation:

As we know by mechanical energy conservation law we have

[tex]mgh = \frac{1}{2}mv^2[/tex]

[tex]v = \sqrt{2gh}[/tex]

since both child starts at same height so here they both will reach the bottom at same speed

3.

Answer:

c. The two accelerations are equal

Explanation:

Since we know that average acceleration of the motion is given as

[tex]a = \frac{v_f - v_i}{\Delta t}[/tex]

since here initial and final speeds are same so they both must have same average acceleration here.