KPK 11th Class Physics Chapter 3 Motion and Force Short Questions Answers

Yes, the velocity of a body can reverse direction even when the acceleration is constant.

For example, a ball thrown in the vertically upwards direction has a constant acceleration ‘g’ which is 9.8 m/sec2. When the body reaches to its maximum height, it reverse its direction and falls vertically downwards towards earth with acceleration g = 9.8 m/sec2. In this way, the velocity of the body reverse its direction while its acceleration due to gravity ‘g’ remains constant.

Both balls will hit the ground with speed. 

Explanation: let the man throws the ist ball from point ‘A’ to ‘B’ at a height ‘h’ from point ‘A’ with initial velocity ‘vi’. Let the ball reaches at point ‘B’ and come to rest and then moves back towards point ‘A’ as shown in the figure. 

Now we know that when a body is thrown upwards with certain velocity ‘vi’ from certain position, then the body will reach that position again with velocity equal to its velocity of projection i.e. its initial velocity ‘vi’. Thus in above case, when the ball reaches point ‘A’ again, its velocity at ‘A’ will be the same to its initial velocity of projection. We have also given that the 2nd ball at ‘A’ is thrown downwards with same speed to that of its ball, so both balls will hit the ground with same speed. The only difference is that both the balls will hit the ground at different times due to different heights as shown in the figure.

Yes, this statement is true when a body moves with constant velocity its acceleration will be zero i.e. a=0 as zero is a constant quantity, so we can say that when a body is moving with constant velocity, the acceleration will remains constant [a = 0 = constant] throughout the motion. Hence it is a special case of motion.

Impulse: The produce of average force and duration of times ‘at’ for which the force acts is known as impulse. It is denoted by J→ and is given by, J = F→ F_one × ∆t _______ (1)

Relation b/w impulse and linear momentum:

According to Newton,s 2nd law, we have F are = ∆p/∆t

=> F are ×∆t = ∆p _______ (2)

Putting equation (2) in equation (1) we get,

J= ∆p ________ (3)

Equation (3) represents the relation b/w impulse and linear momentum. Equation (3) show that impulse is equal to change in linear momentum of a body.

“The collision in which the motion of the objects occurs along a straight line is known as perfect elastic head-on-collision.”

We know that the maximum height of projectile is given by,

  H =   Vi²Sin²θ/2g  …………………(1)

We Also know that 

The  horizontal rang of projectile is given by 

R =  Vi²sin²θ/g

=> R = Vi² 2sinθ cosθ/g  ……………..(3)   [sin²θ=2sinθcosθ]

Dividing and multiplying the R.H.S of equation (3) by sin/2, So we get,

R = Vi² 2sinθcosθ/g × sinθ/2/sinθ/2

=>          R   =  4 cosθ/sinθ [vi²sin²θ/2g]  ……………………………(4)

Putting equation (1) in equation (4), we get,

R = 4 cosθ/sinθ ×H

• R = 4 cot x H
• R = 4 x 1/tanθ x H
• R tan = 4H  ……………………(5)

Now we know that the range of projectile is maximum, it = 45°, so eq (5) because, 

  R_max x tan 45° = 4H

  [ tan 45°=1

• R_max x 1 = 4H
• R_max=4H  ……………………..(6)

Eq (6) shows that for any specific velocity of projection, the range of a projectile cannot exceed from a value equal to four times of the cor-responding height.

According to given conditions, we have. Maximum Height = Horizontal Range

• H = R                                             H=Vi² sin²θ/2g
• Vi²sin²θ/2g    = Vi² sin²/g       R = Vi²  Sin²θ/g
• sin²/2      = 2 sinθ cos
• sinθ = 4 cos                  =>  sinθ/cosθ  = 4 
• tanθ = 4 
• θ = tan-14=75.96°=76°
• θ=76°

Initial speed of object = v_i = 15m/sec,

Final speed of object = v_f=15m/sec

Angle with ‘ox’ = θ=60°

Acceleration of object during the turn = a =?

We know that, 

Slope = tan  ……………(1)

We also know that, 

Slope = Acceleration ………………..(2)

Acceleration = tan

• A = tan60°                        =1.73     m/sec2
• A = 1.73 m/sec²

Aeroplane while horizontally drops a exactly above the target, but missed it. It is because of projectile motion. In this case, the bomb posses both vertical and horizontal components of velocity. Due to horizontal component of velocity, the bomb misses the target and falls slightly ahead the target, as shown in the figure.