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# Tag Archives: 10m

## Conservation of energy Problem

A **ball** is dropped from a **height** of **10m**. If the kinetic **energy** of the **ball** reduces by 40% after striking the **ground**, how high can the **ball** bounce back? (**Elsy James** asked)

## Numerical from centre of mass

Two skaters, one with Mass 45kg, 2’nd with Mass of 65kg stand on an ice rink holding a pole with L=10m with negligible mass. starting from the ends of the pole, the skaters pull themselves along the pole until they meet. How far does the 45kg skater move?

**Hint: **They will meet at the centre of mass of the system.

If the 45 kg skater moves by x m, then the second one travels (10-x)m

then,

45 x = 65(10-x)

Solve and get the value for x

## A simple numerical on momentum

**Car mass is 1200 kg moving velocity is 108 km/hr. On applying brakes velocity reduced to 36 km/hr. Find change in momentum?**

*(posted by Aditi)*

Answer:

m=1200 kg

u=108 x 5/18=30m/s

v=36×5/18=10m/s

Change in momentum = m(v-u) = 1200 x (30-10)=1200 x 20= __24000 kg m/s__

## Numerical Problem from Friction

“The friction coefficient between a road and the tyre of a vehicle is 4/3.find the maximum incline the road may have so that once hard brakes are applied and the wheel starts skidding,the vehicle going down at a speed of 10m/s is stopped within 5m” Raju Rabha asked

## Another problem from kinematics

Zeen asks:

A boy jumps from rest, straight down from the top of a cliff. He falls halfway down to the water below in 0.800 s. How much time passes during his entire trip from the top down to the water? Ignore air resistance.

**Answer **:

let the total height be h

So, **for first case**

S = h/2

a = g=10 m/s^2

u=0 m/s

t = 0.8 sec

using the relation

h/2 = 0.5 x 10×0.8 ^2

h=6.4 m

**In second case (considering the full motion)**

S=h=6.4m

t=?

a=g=10m/s^2

u=o m/s

using the relation

6.4 = 0.5 x 10 x t^2

or

t = 2 x 6.4/10 = **1.28 s**

## A question from kinematics – motion in one dimension

Natasha Sehgal Asked:

“A stone is thrown in a vertically upward direction with a velocity of 5m/s. If the acceleration of the stone during its motion is 10m/s^2 in the downward direction , what will be the height attained by the stone and how much time will it take to reach there?”

Answer:

Take,

u=5m/s

a= -10 m/s^{2}

v=0 at the topmost position

S= H_{max} = ?

The eqn of motion to use

v^{2} – u^{2} = 2aS

Which gives H = 1.25 m

From the eqn

v=u + at

t= 0.5 s