Answer: 1m/s
Explanation: according to the law of conservation of linear momentum in an isolated system, the momentum of the gun equals that of the bullet.
Mathematically
Mb×Vb = Mg×Vg
Where Mb = mass of bullet = 1/100 = 0.01 kg
Vb = velocity of bullet = 200 m/s
Mg = mass of gun = 2kg
Vg = recoil velocity of gun =?
0.01×200 = 2×Vg
Vg = 0.01×200/2
Vg = 0.01×100
Vg = 1m/s
Answer: 29.50 m
Explanation: In order to calculate the higher accelation to stop a train without moving the crates inside the wagon which is traveling at constat speed we have to use the second Newton law so that:
f=μ*N the friction force is equal to coefficient of static friction multiply the normal force (m*g).
f=m.a=μ*N= m*a= μ*m*g= m*a
then
a=μ*g=0.32*9.8m/s^2= 3.14 m/s^2
With this value we can determine the short distance to stop the train
as follows:
x= vo*t- (a/2)* t^2
Vf=0= vo-a*t then t=vo/a
Finally; x=vo*vo/a-a/2*(vo/a)^2=vo^2/2a= (49*1000/3600)^2/(2*3.14)=29.50 m
The correct answer would be the last one
Answer:32 m/s/s
Explanation: since F=M*A, F=16N, M=0.5kg, A= F/M
A=16/0.5
A=32 m/s/s
