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101.

A `10g` bullet is short from a `5kg` gun with a velocity of `400m//s`. What is the speed of recoil of the gun?

Answer» Here, mass of bullet, `m_(1)= 10g= 10xx10^(-3)kg= 10^(2)kg`, mass of gun, `m_(2)= 5kg`
velocity of bullet, `v_(1) = 400m//s`, speed of recoil of gun `v_(2) = ?`
As total momentum of bullet and gun after firing = total momentum before firing, `m_(1)v_(1)+m_(2)v_(2)=0`
or `v_(2) = -(m_(1)v_(1))/m_(2)= -(10^(-2)xx400)/5` or `v_(2)= -0.8m//s`
Negative sign shows that the gun moves in a direction opposite to that of the bullet.
Discussion
102.

Velocity of recoil of gun is much smaller than the velocity of bullet. This is beaacuseA. gun is much heavier than the bulletB. gun is held tightlyC. either (a) and (b)D. neither (a) or (b).

Answer» Correct Answer - A
As `m_(1)v_(1)+m_(2)v_(2)=0 :. v_(2)= - (m_(1)v_(1))/(m_(2))`
As `m_(2) gt gt m_(1)`, therefore `(-)v_(2)lt lt v_(1)`
Discussion
103.

Recoil velocity of gun isA. equal to velocity of bulletB. much greater than velocity of bulletC. much smaller than velocity of bulletD. can not say

Answer» Recoil velocity of a gun is much smaller than the velocity of bullet.
Discussion
104.

Recoil velocity of gun is much smaller than the velocity of bullet. Why?

Answer» This is because the gun is far more heavier than the bullet.
Discussion
105.

A car and a truck moving with the same velocity in opposite directions collide and collision lasts for `2 s`. (i) Which vehicle experiences the greater force of impact? (ii) Which vehicle has the smaller change in momentum? (iii) Which one undergoes greater acceleration? (iv) Which one is pushed farther? (v) Which one is likely to suffer more damage?

Answer» Let mass of truck `=M`, mass of car `=m`
velocity of truck `=v`
velocity of car `= -v` (Negative sign for opposite direction of motion)
time for which collision lasts, `t=1s`
(a) On collision, both the vehicles experience the same force, as action and reaction are equal.
(b) Change in momentum of truck is equal and opposite to change in momentum of car, i.e., both the vehicles experience the same change in momentum.
(c) As acceleration=force/mass, and force on each vehicle in same, therefore, acceleration `prop1/("mass")`
As mass of car is smaller, therefore, acceleration of car is greater than the acceleration of the truck.
(d) The car is likely to suffer more damage than the truck, as it is lighter. The acceleration, i.e., change in velocity of car is more than of the truck.
Discussion
106.

The speed- time graph of a car is given in (figure) The car weights `1000kg` (a) What is the distance travelled by the car in first two seconds? (b) What is the braking force applied at the end of `5 seconds` to bring the car to a stop within one second?

Answer» (a) Here, mass of car,
`m=1000kg` distance, `s=?`
In the first two seconds, the speed- time graph is `OA`
s=area of `DeltaOAE= ("base" xx "height")/2= (OExxEA)/2` or `s=(2xx15)/2= 15m`
(b) initial speed, `u=15m//s` final speed, `v=0`
time taken, `t=1s` Barking force, `F=?`
As `F= ma= (m(v-u))/t,F= (1000(0-15))/1= -15000N`
Negative sign is for opposing force.
Discussion
107.

A truck of mass `2500 kg` moving at `15m//s` collides with a car of mass `1000kg` moving at `5m//s` in the opposite direction . With what velocity would the two move together?

Answer» Here, `m_(1)= 2500 kg , u_(1)= 15m//s, m_(2)=1000kg , u_(2)= -5m//s`
Negative sign with `u_(2)` indicates that `u_(2)` is in a direction opposite to `u_(1)`.
If `v` is the combined velocity of the two, then applying the principle of conservation of linear momentum, we get `m_(1)u_(1)+m_(2)u_(2)=(m_(1)+m_(2))v` or `2500xx15+1000(-5)=(2500+1000)v`
or `37500-5000= 3500v` or `v=(32500)/(3500)= 65/7= 9.3m//s`
Positive sign of `v` indiacates that the resultant velocity is in the direction of motion of the truck.
Discussion
108.

A truck of mass `M` is moved under a force `F`. If the truck is then loaded with an object equal to the mass of the truck and the driving force is halved, then how does the acceleration change?

Answer» As mass of loaded truck becomes 2 times and driving force is halved, therefore, from
`F=ma, a=F/m=(1//2 "time")/(2 "time")=1/4` time
i.e., acceleration would become `1/4th`
Discussion
109.

A hockey ball of mass `200g` travelling from west to east at `10m//s` is struck by a hockey stick. As a result, the ball gets turned back and now has a speed of `5m//s`. If the ball and hockey stick were in contact for `0.2s`, Calculate (i) initial and final momentum of the ball (ii) rate of change of momentum of the balll (iii) force exerted by hockey stick on the ball.

Answer» Here, `m=200g = 0.2kg, u=10m//s` west to east, `v= -5m//s`, east to west, `t=0.2s`.
(i) Initial momentum of ball `=m u = 0.2xx10= 2kg m//s` , from west to east.
Final momentum of the ball `=mv=0.2(-5)= -1 kgm//s` , from east to west.
(ii) Rate of change of momentum of ball `=("final momentum -i nitial momentum")/("time taken")=(-1-2)/(0.2)`
(iii) Force exerted by hockey stick on the ball = rate of change of momentum of ball
`= -15kg m//s^(2)= -15N` from east to west.
Discussion
110.

Two balls of the same size but different matetials, rubber and iron are kept on the smooth floor of a moving train. The brakers are applied suddenly to stop the train . Will the balls start rolling? If so, in which direction? Will they move with the same speed? Give reasons for your answers.

Answer» Yes, both the balls will start rolling due to inertia of motion. The train stops due to brakes. But the balls continue to move in the direction of the train. The two balls will not roll with the same speed. Infact, the iron ball being heavier than the rubber ball shall move at a lower speed.
Discussion
111.

A water tanker filled up to `2/3` of its height is moving with a uniform speed. On sudden application of the brake, the water in the tank wouldA. move backwardB. move forwardC. be unaffectedD. rise upwards

Answer» Correct Answer - B
On sudden application of brakes, water in the tank would move forward due to inertia of motion.
Discussion
112.

Two identical bullets are fired one by a light rifle and another by a heavy rifle with the same force. Which rifle will hurt the shoulder and why?

Answer» As momentum of light rifle and heavy rifle in opposite direction is same (being equal to momentum of identical bullets), the lighter rifle will move faster hurting the shoulder more.
Discussion
113.

Using a horizontal force ` 200N`, we intend to move a wooden cabinet across a floor at constant velocity. What is the frictional force that will be exerted on the cabinet?

Answer» Force applied, `P=200N`, force of friction, `F=?`
As the wooden cabinet is to move across the floor with a constant velocity, no force `(f)` is spent in accelerating the cabinet, i.e., `f=P-F= 0` or `F=P = 200N`
Discussion