In previous exercise we solved the exercise which includes questions of interia Intext Exercise 1.
In this exercise we will solve questions related to the law of motion and momentum.
1. If action is always equal to the
reaction, explain how a horse can
pull a cart.
Answer:
When the horse pushes the earth with foot while moving forward according to NEWTON’S THIRD Law the earth pushes back and the cart moves forward.
2. Explain, why is it difficult for a
fireman to hold a hose, which
ejects large amounts of water at
a high velocity.
Answer: When fireman holds the hose a force in backward direction is applied due to ejection of water.
Due to ejection of large amount of water in high velocity, very large backward force Is applied( in reaction of forward ejection) due to this force the fireman holding hose is imbalances.
3. From a rifle of mass 4 kg, a bullet
of mass 50 g is fired with an
initial velocity of 35 m s–1.
Calculate the initial recoil velocity
of the rifle.
Answer :
Given : mass of rifle, m1= 4kg
Mass of bullet, m2= 50g
Initial velocity of system v =0
Recoil velocity of gun, v1 =?
Initial velocity of bullet, v2 = 35m/s
So momentum before firing
(m1+m2) v = 0
Momentum after firing
=m1v1+m2v2
= 4*v1+ 0.050*35
Momentum after the motion = momentum before the motion
4v1 + 1.75 =0
v1 = -0.435 m /s
the negative sign shows the backward direction of recoil velocity.
4. Two objects of masses 100 g and
200 g are moving along the same
line and direction with velocities
of 2 m s–1 and 1 m s–1, respectively.They collide and after the collision,
the first object moves at a velocity
of 1.67 m s–1. Determine the
velocity of the second object.
Answer :
Given : mass of object 1,m1= 100g
mass of object 2, m2 = 200g
Before collision
Velocity of object 1,v1= 2m/s
Velocity of object 2,v2 = 1m/s
After collision
The velocity of object 1 v1′ = 1.67m/s
Momemtum before collision
m1v1+m2v2= 0.2*2+0.1*1
= 0.4+0.1
= 0.5
After collision
m1v1’+ m2v2’= 0.2*1.67+ 0.1*v2′
= 0.334 + 0.1v2′
Momemtum will be conserved
0.5 = 0.334 + 0.1v2′
0.5 – 0.334 = 0.1v2′
0.166 = 0.1 v2′
0.0166 = v2′
Hope this will help you we will solve the next exercise in next post.
