Newton S Laws Of Motion(11th And 12th > Physics ) Questions and Answers
Explanation:-
Answer: Option C. -> 40 km/hour:
C
Initially the wagon of mass 1000 kg is moving with velocity of 50 km/h
So its momentum =1000×50kg×kmh
When a mass 250kg is dropped into it. New mass of the system = 1000+250 = 1250 kg
Let v is the velocity of the system.
By the conservation of linear momentum : Initial momentum = Final momentum
1000×50=1250×v
v=50,0001250=40kmh.
Two blocks are in contact on a frictionless table one has a mass m and the other 2 m as shown in figure. Force F is applied on mass 2m then system moves towards right. Now the same force F is applied on m. The ratio of force of contact between the two blocks will be in the two cases respectively.
Explanation:-
Answer: Option B. -> 1 : 2:
B
When the force is applied on mass 2m contact force f1=mm+2mg=g3
When the force is applied on mass m contact force f2=2mm+2mg=23g
Ratio of contact forces f1f2=12
Explanation:-
Answer: Option C. -> 8N:
C
By comparing the above problem with general expression. T1=(m2+m3)m1+m2+m2=(3+5)102+3+5=8 Newton
Three masses of 15 kg. 10 kg and 5 kg are suspended vertically as shown in the fig. If the string attached to the support breaks and the system falls freely, what will be the tension in the string between 10 kg and 5 kg masses. Take g=10ms−2 . It is assumed that the string remains tight during the motion
Explanation:-
Answer: Option D. -> Zero:
D
In the condition of free fall, tension becomes zero.