A bullet of mass 0.01 kg is moving horizontally with a speed of 100 m/s when it hits a block of mass 2 kg that is at rest on a horizontal surface with a coefficient of friction of 0.4. After the collision the bullet becomes embedded in the block.
Wow, that's quite an action-packed scenario you've got there! I can already smell the blockbuster movie adaptation. But let me put on my Comically Over-Exaggerating Scientific Expert hat and try to help you with the aftermath.
First, let's calculate the initial momentum of the bullet before the collision:
Momentum (P) = mass (m) × velocity (v)
P = 0.01 kg × 100 m/s
P = 1 kg·m/s
Now, after the bullet embeds itself in the block, we can treat the bullet-block system as a single object. To conserve momentum, we assume there is no external force acting on it. Therefore, the momentum after the collision should still be 1 kg·m/s.
Next up, the friction between the block and the surface. The coefficient of friction is given as 0.4, which means the frictional force (F) can be calculated using:
F = coefficient of friction (μ) × normal force (N)
The normal force (N) is equal to the weight of the block, which is:
N = mass (m) × gravity (g)
N = 2 kg × 9.8 m/s²
N = 19.6 N
Plugging in the values, we find:
F = 0.4 × 19.6 N
F = 7.84 N
Now, since the block is at rest and we have a frictional force acting on it, we can say that this force equals the maximum static frictional force (Fs). This handy force depends on the coefficient of static friction (μs) and the normal force (N).
Fs = μs × N
Given that the static friction coefficient is the same as the kinetic friction coefficient (0.4), we know that Fs equals 7.84 N.
So what happens next? We compare the frictional force acting on the block (7.84 N) to the force required to slide the block (static friction). If the bullet-block system's momentum is enough to overcome the static friction, the block will start moving.
If the momentum is not enough, well, then we have ourselves a standstill situation worthy of a circus act. The clown cars will start lining up, popcorn will be consumed, and spectators will enjoy the show while waiting for some external force to break the static friction and send that block flying.
Hope this response gave you a good laugh while explaining the physics involved! Remember, even in scientific matters, there's always room for a little clowning around.
To solve this problem, we need to consider the conservation of momentum and the laws of motion.
Step 1: Calculate the initial momentum of the bullet.
Momentum (p) is given by the formula p = mass × velocity.
The initial momentum of the bullet (p_i) is 0.01 kg × 100 m/s = 1 kg·m/s.
Step 2: Calculate the momentum of the block.
Since the block is initially at rest, its initial momentum (p_i) is 0 kg·m/s.
Step 3: Calculate the total momentum before the collision.
The total initial momentum (p_i_total) before the collision is the sum of the momenta of the bullet and block: p_i_total = p_i_bullet + p_i_block = 1 kg·m/s + 0 kg·m/s = 1 kg·m/s.
Step 4: Determine the final velocity of the embedded bullet-block system.
According to the law of conservation of momentum, the total momentum before the collision is equal to the total momentum after the collision.
So, we have: p_f_total = p_f_bullet + p_f_block = p_i_total.
Step 5: Calculate the final velocity of the embedded bullet-block system.
The final momentum of the block (p_f_block) is given by the formula p_f_block = mass_block × velocity_block.
Since the bullet becomes embedded in the block, the final momentum of the bullet (p_f_bullet) will be equal to the mass of the bullet plus the mass of the block combination, multiplied by their common final velocity (v_final): p_f_bullet = (mass_bullet + mass_block) × v_final.
Step 6: Apply the coefficient of friction to find the final velocity.
The friction force between the block and the horizontal surface will work against the motion, so we can use the equation of motion:
Force of friction = (coefficient of friction) × (normal force).
The normal force is equal to the weight of the block, which is given by the formula: weight_block = mass_block × acceleration_due_to_gravity.
The friction force can be calculated as: friction force = (coefficient of friction) × (normal force).
Using Newton's second law (F = m × a), we can calculate the acceleration of the block during the collision: acceleration_block = friction force / mass_block.
Since the block is initially at rest, we can find the time taken to come to rest (t) using the equation of motion: 0 = v_final - (acceleration_block × t).
Step 7: Final calculation.
Substituting the value of t from Step 6 into our equation from Step 5, we can find the final velocity (v_final):
0 = v_final - (acceleration_block × t)
v_final = acceleration_block × t.
Now, you can plug in the values given in the problem and calculate the final velocity.
To find the final velocity of the combined block and bullet system after the collision, we can apply the principle of conservation of momentum.
Step 1: Find the initial momentum of the bullet and block system before the collision.
The momentum (p) of an object is calculated by multiplying its mass (m) by its velocity (v): p = m * v.
For the bullet:
Mass of the bullet, m1 = 0.01 kg
Initial velocity of the bullet, v1 = 100 m/s
Momentum of the bullet, p1 = m1 * v1 = 0.01 kg * 100 m/s = 1 kg*m/s
For the block:
Mass of the block, m2 = 2 kg
The block is at rest, so the initial velocity of the block, v2 = 0 m/s
Momentum of the block, p2 = m2 * v2 = 2 kg * 0 m/s = 0 kg*m/s
The initial momentum of the bullet and block system, p_initial = p1 + p2 = 1 kg*m/s + 0 kg*m/s = 1 kg*m/s
Step 2: Find the final momentum of the combined block and bullet system after the collision.
Since the bullet becomes embedded in the block after the collision, their masses combine to form a new mass (m3) which is the sum of their individual masses.
Mass of the combined system, m3 = m1 + m2 = 0.01 kg + 2 kg = 2.01 kg
Let the final velocity of the combined system after the collision be v_final (which we need to find).
Final momentum of the combined system, p_final = m3 * v_final
According to the principle of conservation of momentum, the initial momentum (p_initial) should be equal to the final momentum (p_final) of the system.
Therefore, p_initial = p_final
1 kg*m/s = 2.01 kg * v_final
Now we can solve for v_final:
v_final = 1 kg*m/s / 2.01 kg ≈ 0.4975 m/s
Therefore, the final velocity of the combined block and bullet system after the collision, where the bullet is embedded in the block, is approximately 0.4975 m/s.
initial momentum = 0.01 * 100
momentum immediately after collision = 2.01 Vi
so
2.01 Vi = 0.01 * 100
solve for Vi
Then you have a force of F = -2.01 * g * 0.4 retarding the motion
a = F/m = F/2.01 = -0.4 g
V = Vi -0.4 g t
x = Vi t -(0.4 g/2)t^2