Calculations Of Motion Along A Straight Line

Question 1

Consider an object released from rest,falling, hitting a bed of sand. Describe the motion in two stages.

Answer 1

1. Falling motion due to gravity: acceleration= g (downwards)
2. Deceleration in the sand: initial velocity= velocity of object just before impact

Question 2

What is the link between the two stages of an object being released from rest, falling, then hitting a bed of sand

Answer 2

The acceleration is not the same in each stage. However, the value that stays the same is for the velocity at the end of the first stage and the velocity at the start of the second stage.

Question 3

A ball is released from a height of 0.85m above a bed of sand and creates an impression of depth 0.025m. Calculate the two stages of motion.

Answer 3

Stage 1: u=0s, s=-0.85m, a= -9.8 ms^-2

v^2= u^2 +2as

v^2= 0^2 + 2 x -9.8 x -0.85
=16.7m^2s^-2
= -4.1ms^-1

Stage 2: u= -4.1ms^-1, v=0, s=-0.025m

v^2= u^2 +2as

0^2 =(-4.1)^2 + 2a x -0.025
2a x 0.025=16.8

a= 16.8/ (2 x 0.025)
= 336 ms^-2