Question

The surface is tilted to an angle of 37 degrees from the horizontal, as shown above in Figure 3. The blocks are each given a push so that at the instant shown, they are moving toward each other. ii. At the instant shown in Figure 3, the blocks are moving toward each other with the same speed of 0.35m/s . The blocks collide 0.50 seconds later. What is the speed of the two-block system’s center of mass just before the blocks collide?

Answer 1

The velocity of the center of mass of the two-block system just before the collision is 0 m/s, as they are moving at the same speed toward each other and have the same mass, resulting in no net movement of the center of mass.

Step-by-step explanation:

To determine the speed of the two-block system's center of mass just before the collision, we can use the conservation of momentum. Assuming no external forces are acting on the system along the direction of the blocks' movement, the speed of the center of mass will not change due to the collision. Since both blocks are moving toward each other with the same speed of 0.35 m/s and presumably have the same mass (as the problem doesn't provide differing masses), the center of mass of the system is at rest relative to an observer on the surface.

The velocity of the center of mass (Vcm) before the collision can be calculated using the formula: Vcm = (m1*v1 + m2*v2) / (m1 + m2), where m1 and m2 are the masses of the blocks and v1 and v2 are their respective velocities. Here, we can see that v1 is +0.35 m/s and v2 is -0.35 m/s, which makes the overall momentum of the system zero. Therefore, the velocity of the center of mass just before the blocks collide is 0 m/s.

Answer 2

Incomplete question: "Each block has a mass of 0.2 kg"

The speed of the two-block system's center of mass just before the blocks collide is 2.9489 m/s

Step-by-step explanation:

Given data:

θ = angle of the surface = 37°

m = mass of each block = 0.2 kg

v = speed = 0.35 m/s

t = time to collision = 0.5 s

Question: What is the speed of the two-block system's center of mass just before the blocks collide, vf = ?

Change in momentum:

`delta(P)=F*delta(t)`

`P_(f) -P_(i)=F*delta(t)`

`2m(v_(f) -v_(i))=F*delta(t)`

`v_(i) =0.35-0.35=0`

It is neccesary calculate the force:

`F=(m+m)*g*sin\theta`

Here, g = gravity = 9.8 m/s²

`F=(0.2+0.2)*9.8*sin37=2.3591N`

`v_(f) =(F*delta(t))/(2m) =(2.3591*0.5)/(2*0.2) =2.9489m/s`