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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?

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

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.

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 $S=ut + \frac{1}{2}at^{2}$

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 $S=ut + \frac{1}{2}at^{2}$
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

“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?”

Take,

u=5m/s

a= -10 m/s2

v=0 at the topmost position

S= Hmax = ?

The eqn of motion to use
v2 – u2 = 2aS
Which gives H = 1.25 m
From the eqn
v=u + at
t= 0.5 s

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