A 10.00-g bullet traveling horizontally at 533 m/s hits a1.00-kg wooden block, which is initially at rest on a rough horizontal surface. The coefficient of kinetic friction between the block and the surface is 0.4. The bullet buries itself in the block without passing through. The distance the block moved on the surface before stopping is:Question 12Select one:a.4.48 mb.6.89 mc.5.86 md.3.45 me.2.41 m
Question
A 10.00-g bullet traveling horizontally at 533 m/s hits a1.00-kg wooden block, which is initially at rest on a rough horizontal surface. The coefficient of kinetic friction between the block and the surface is 0.4. The bullet buries itself in the block without passing through. The distance the block moved on the surface before stopping is:Question 12Select one:a.4.48 mb.6.89 mc.5.86 md.3.45 me.2.41 m
Solution
To solve this problem, we need to use the principles of conservation of momentum and energy.
Step 1: Calculate the initial momentum of the system. The bullet has momentum (mass x velocity) and the block is at rest so its momentum is zero. The total initial momentum is therefore (10 g x 533 m/s) = 5330 kg.m/s.
Step 2: After the bullet hits the block, they move together. So, their combined mass is (10 g + 1 kg) = 1.01 kg. The final momentum of the system is also 5330 kg.m/s (since momentum is conserved). So, we can find the final velocity (v) of the block and bullet using the equation for momentum (mass x velocity): 1.01 kg x v = 5330 kg.m/s. Solving for v gives v = 5273 m/s.
Step 3: The block and bullet then slide and stop due to friction. The work done by friction (which is equal to the kinetic energy of the block and bullet) can be calculated using the equation for work (force x distance). The force of friction is the mass of the block and bullet times gravity times the coefficient of friction (1.01 kg x 9.8 m/s^2 x 0.4 = 3.96 N). The kinetic energy of the block and bullet is 1/2 times mass times velocity squared (0.5 x 1.01 kg x (5273 m/s)^2 = 14013 J).
Step 4: Setting the work done by friction equal to the kinetic energy gives: 3.96 N x distance = 14013 J. Solving for distance gives distance = 14013 J / 3.96 N = 3540 m.
So, none of the options (a.4.48 m, b.6.89 m, c.5.86 m, d.3.45 m, e.2.41 m) are correct. The block moves 3540 m before stopping.
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