A ball is travelling with uniform translatory motion. This means that
it is at rest
the path can be a straight line or circular and the ball travels with uniform speed
all parts of the ball have the same velocity (magnitude and direction) and the velocity is constant
the centre of the ball moves with constant velocity and the ball spins about its centre uniformly
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In a uniform translatory motion if all the parts of the body moves with (same velocity in same straight line, so the velocity is constant.
The situation is shown in (figure) where a body is in unfirom translatory motion.
A metre scale is moving with uniform velocity. This implies.
the force acting on the scale is zero, but a torque about the centre of mass can act on the scale
the force acting on the scale is zero and the torque acting about centre of mass of the scale is also zero
the total force acting on it need not be zero but the torque on it is zero
neither the force nor the torque need to be zero
According to question we have to apply Newton's second law of motion, in terms of force and change in momentum.
We know that
As given that the meter scale is moving with uniform velocity, hence
Force
No change in its velocity i.e., acceleration of it zero by
Newton's second law.
Hence, net or resultant force must act on body is zero:
, so,
As all part of the scale is moving with uniform velocity and total force is zero, hence, torque will also be zero.
A cricket ball of mass has an initial velocity and a final velocity , after being hit. The change in momentum (final momentuminitial momentum) is (in )
zero
As given that,
Mass of the ball
Change in momentum
Final momentum - Initial momentum
Hence verifies option (c).
In the previous problem (3), the magnitude of the momentum transferred during the hit is
From previous solution
Magnitude of
Verifies the option (c).
Conservation of momentum in a collision between particles can be understood from
Conservation of energy
Newton's first law only
Newton's second law only
both Newton's second and third law
From the FBD
and
A hockey player is moving northward and suddenly turns westward with the same speed to avoid an opponent. The force that acts on the player is
frictional force along westward
muscle force along southward
frictional force along sotuh-west
muscle force along south-west
Consider the adjacent diagram.
The force on player is due to rate of change of momentum.
The direction of force acting on player will be the same as the direction of change in momentum.
Let i.e., Initial momentum of player northward i.e., Final momentum of player towards west.
Clearly,
= Clearly resultant will be along south-west.
So, it will be also the direction of force on player.
A body of mass travels according to the law where, and . The force acting on the body at is
As given that, mass
As given equation is
Force acting on body .
A body with mass is acted upon by a force . If its initial velocity at is , the time at which it will just have a velocity along the -axis is
never
As given that mass
Acting force
Initial velocity at
Retardation,
As the final velocity along -component only. So its component must be zero.
From , for -component only,
. Hence verifies the option (b).
A car of mass starts from rest and acquires a velocity along east, in two seconds. Assuming the car moves with uniform acceleration, the force exerted on the car is
eastward and is exerted by the car engine
eastward and is due to the friction on the tyres exerted by the road
more than eastward exerted due to the engine and overcomes the friction of the road
exerted by the engine
As given that mass of the car
As car starts from rest,
Velocity acquired along east
Time .
We know that,
(Force, by engine is internal force)
Hence, force acting on the car is towards east due to force of friction is which moves the car in eastward direction. Hence, force by engine is internal force.
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