Laws of Motion - Online Test

Answer : Option D
Explaination / Solution:

When the lift fall freely under gravity, a = g

Therefore Apparent weight, (We experience weight due to reaction)

R=m(g−a)=m(g−g)=0

This is the condition for weightlessness.

Answer : Option B
Explaination / Solution:

When the lift falls freely under gravity, a=g

Therefore Apparent weight, 

R=m(g−a)=m(g−g)=0

This is the condition of weightlessness.

Answer : Option C
Explaination / Solution:

When t < 0. As this part is horizontal, thus it can be concluded that distance covered by the particle is zero and hence force on the particle is zero. When 0 < t < 4 s. As OA has a constant slope, hence in this interval, particle moves with constant velocity (acceleration = 0). Hence force on the particle is zero. W'hen t > 4s. As this portion shows that particle always remains at a distance of 3 m from the origin i.e., the particle is at rest. Hence force on the particle is zero.

Answer : Option D
Explaination / Solution:

Initial velocity u=0$u=0$, between t= 0s and t=4s, the particle has a constant velocity,

therefore slope of OA is  v=34m/s

At t= 0,Impulse=Change in momentum

                        = m(v−u)=4(34−0)=3kgm/s

At t= 4s, initial velocity is 3/4 m/s,

So, the Impulse = m(v−u)=4(0−34)=−3kgm/s

Answer : Option D
Explaination / Solution:

Let,

T =tension in the string,

a= acceleration in the direction of force F. 

m1=10 kg and m2= 20 kg

Therefore,common acceleration a=Fm1+m2=60010+20=20ms−2

Now the tension T in the string is T=m1a=10×20=200N

Answer : Option C
Explaination / Solution:

Let, T= tension on the string

a = acceleration, m1 = mass of light body= 10 kg and m2= mass of heavy body = 20 kg

a=Fm1+m2=60030=20ms−2

Now the tension in the string will be,

T=m2a=20×20=400N

Answer : Option B
Explaination / Solution:

Let, M = mass of nucleus at rest . m1 and m2 are masses of two smaller nuclei. v1 and v2 are the velocities of respective masses

Now, According to the law of conservation of momentum,

Initial momentum before disintegration = final momentum after disintegration

m1v1+m2v2=0

v2=−m2m1⋅v1

As masses m1$m_1$and m2$m_2$ cannot be negative, v1 and v2 having opposite signs and so the two smaller nuclei move in opposite directions. 

Answer : Option D
Explaination / Solution:

Impulse imparted to one ball by the other = change in momentum

= momentum after collision - momentum before collision

= {0.05 x 6} - {0.05x(-6)} 

=  0.6 kg m s-1 

Answer : Option B
Explaination / Solution:

Massof shell, m=0.02kg

Mass of gun, M=100kg

Speed of shell, v=80ms−1

Let V be the recoil speed of the gun. According to the law of conservation of momentum,

Initial momentum = Final momentum

0=mv+MVV=−mvM=−0.02×80100V=−0.016ms−1=−1.6cms−1

Negative sign indicates that the gun moves backward as the bullet moves forward.

Answer : Option A
Explaination / Solution:

Let, m=0.25 kg, r= 1.5m and f= 40 rev/min = 40 rev /60 s= 2/3 rps

ω=2πf=2π×23=4π3rads−1

Tension in the  string = Centripetal force

so, T=mrω2=0.25×1.5×(4π3)2=6.6N