## To determine the speed of block 3, we can use the principle of conservation of momentum. The initial momentum of block 1 is given as (m1)(v1i), where m1 is the mass of block 1 and v1i is its initial speed.

After the collision with block 2, the total momentum will be (v)(m1 + 0.5m1), where v is the speed of block 2 after the collision. Since block 2 is stationary initially, its velocity is 0.

After the collision with block 3, the total momentum will be (u)(1.5m1 + 0.25m1), where u is the speed of block 3. We know that m3 = 0.5m2 = 0.5(0.5m1) = 0.25m1.

Setting the initial momentum equal to the final momentum, we have (m1)(v1i) = (u)(1.75m1).

Solving for u, we find that u = 2.3 m/s.

For part (b), we can compare the kinetic energy (KE) of block 3 to the KE of block 1. The KE of block 3 is given by 1/2(mu^2), where m is the mass of block 3 and u is its speed. The KE of block 1 is given by 1/2(m1)(v1i^2). We can calculate these values using the known velocities.

Finally, for part (d), we compare the momentum of block 3 to the momentum of block 1. The momentum of block 3 is given by (0.25m1)(u), while the momentum of block 1 is given by (m1)(v1i). We can compare these values to see if they are greater than, less than, or the same as the initial momentum of block 1.

It seems there was confusion in the previous response about the second calculation. The correct calculation should show that block 3 has a higher velocity than block 1. This can be demonstrated with experiments, such as dropping two balls of different sizes and observing the smaller one bouncing higher.