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(a) In Problem 8, using energy techniques rather than the techniques of Chapter 4, find the speed of the snowball as it reaches the ground below the cliff. What is that speed

(b) If the launch angle is changed to 41.0° belowthe horizontal and

(c) If the mass is changed to 2.50kg?

Short Answer

Expert verified

(a) The speed of the snowball as it reaches the ground below the cliff is 21 m/s.

(b) The speed if the launch angle is changed to 41.00below the horizontal is 21 m/s.

(c) The speed if the mass is changed to 2.50 kg is 21 m/s.

Step by step solution

01

Step 1: Given data

vi=14.0m/s

h=12.5m

02

Determining the concept

The problem is based on the law of conservation of energy which states that the total energy of an isolated system remains constant.According tothe conservation of momentum, momentum of a system is constant if no external forces are acting on the system.

Formula:

Initialtotalenergy=Finaltotalenergy

03

(a) Determining the speed of the snowball as it reaches the ground below the cliff

According to the law of conservation of energy,

KEi+PEi=KEf+PEf

12mvi2+mgh=12mvf2+0

Hence,

vf=vi2+2gh=142+2×9.8×12.5=21m/s

Hence, the speed of the snowball as it reaches the ground below the cliff is 21 m/s.

04

(b) Determining the speed if the launch angle is changed to 41.0°below the horizontal

Here, final velocity does not change.

Hence,

vf=21m/s

Hence,the speed if the launch angle is changed to 41.0obelow the horizontal is 21 m/s.

05

(c) Determining the speed if the mass is changed to  2.50 kg

Velocity found in part (a) doesn’t depend on mass. So, it would bethe same.

vf=21m/s

Hence, the speed if the mass is changed to 2.50 kg is 21 m/s.

Therefore, the unknown speed can be found by using the law of conservation of energy.

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