Answer:
261.3 m/s
Explanation:
Mass of bullet=m=15 g=
1 kg=1000g
Mass of block=M=3 kg
d=0.086 m
Total mass =M+m=3+0.015=3.015 kg
K.E at the time strike=Gravitational potential energy at the end of swing
Using g=
Substitute the values
Velocity after collision=V=1.3 m/s
Velocity of block=v'=0
Using conservation law of momentum
Using the formula
Answer:
v₂=- 34 .85 m/s
v₁=0.14 m/s
Explanation:
Given that
m₁=70 kg ,u₁=0 m/s
m₂=0.15 kg ,u₂=35 m/s
Given that collision is elastic .We know that for elastic collision
Lets take their final speed is v₁ and v₂
From momentum conservation
m₁u₁+m₂u₂=m₁v₁+m₂v₂
70 x 0+ 0.15 x 35 = 70 x v₁ + 0.15 x v₂
70 x v₁ + 0.15 x v₂=5.25 --------1
v₂-v₁=u₁-u₂ ( e= 1)
v₂-v₁ = -35 --------2
By solving above equations
v₂=- 34 .85 m/s
v₁=0.14 m/s
Answer:
Explanation:
It says “Momentum before the collision is equal to momentum after the collision.” Elastic Collision formula is applied to calculate the mass or velocity of the elastic bodies.
Answer:
-0.0789 m
Explanation:
Recall that the y-component comes associated with the sin(18.4) through the following trigonometric relationship:
y = 0.250 sin(-18.4) ≈ -0.0789 m
Notice it is negative since it is below the x-axis.
Answer:
In ideal case, when no resistive forces are present then both the balls will reach the ground simultaneously. This is because acceleration due to gravity is independent of mass of the falling object. i.e. g = GM/R² where G = 6.67×10²³ Nm²/kg², M = mass of earth and R is radius of earth.
Let us assume that both are metallic balls. In such case, we have to take into account the magnetic field of earth (which will give rise to eddy currents, and these eddy currents will be more, if surface area will be more) and viscous drag of air ( viscous drag is proportional to radius of falling ball), then bigger ball will take slightly more time than the smaller ball.
Explanation:
In ideal case, when no resistive forces are present then both the balls will reach the ground simultaneously. This is because acceleration due to gravity is independent of mass of the falling object. i.e. g = GM/R² where G = 6.67×10²³ Nm²/kg², M = mass of earth and R is radius of earth.
Let us assume that both are metallic balls. In such case, we have to take into account the magnetic field of earth (which will give rise to eddy currents, and these eddy currents will be more, if surface area will be more) and viscous drag of air ( viscous drag is proportional to radius of falling ball), then bigger ball will take slightly more time than the smaller ball.