The man fires an 75g arrow so that it is moving at 80 m/s when it hits and embeds in a 13kg block resting on ice. What is the velocity of the block and arrow just after the collision? How far will the block slide on the ice before stopping? A 7.2N friction force opposes its motion.
M!*V1 + M2*V2 = M1*V + M2*V
0.075*80 + 13*0 = 0.075*V + 13*V
6 + 0 = 13.075V
Momentum before the collision=Momentum
after the collision:
M1*V1 + M2*V2 = M1*V + M2*V
0.075*80 + 13*0 = 0.075*V + 13*V
6 + 0 = 13.075V
V = 0.459 m/s
To solve this problem, we can apply the principles of conservation of momentum and conservation of energy.
1. Velocity of the block and arrow just after the collision:
- Conservation of momentum states that the total momentum before the collision is equal to the total momentum after the collision.
- Let's define the initial velocity of the block as v_b and the initial velocity of the arrow as v_a.
- The mass of the arrow is given as 75g, which is 0.075kg.
- According to the problem, the arrow's initial velocity, v_a, is 80 m/s.
- The mass of the block is given as 13kg.
- Let's assume the final velocity of the arrow and block after the collision is v_f.
- Using the principle of conservation of momentum, we can write the equation:
(mass of arrow * velocity of arrow before collision) + (mass of block * velocity of block before collision) = (mass of arrow * velocity of arrow after collision) + (mass of block * velocity of block after collision)
(0.075kg * 80 m/s) + (13kg * 0 m/s) = (0.075kg * v_f) + (13kg * v_f)
- Simplifying the equation:
6kg m/s = (0.075kg + 13kg) * v_f
v_f = 6kg m/s / 13.075kg
v_f ≈ 0.459 m/s
Therefore, the velocity of the block and arrow just after the collision is approximately 0.459 m/s.
2. Distance the block slides on the ice before stopping:
- Conservation of energy can be applied here.
- The work done by the friction force (W_friction) is equal to the force of friction multiplied by the distance the block slides.
- The initial kinetic energy of the block is given by (1/2) * mass * (initial velocity)^2.
- The final kinetic energy of the block is zero because it comes to a stop.
- The work done by friction force, which opposes motion, is equal to the change in kinetic energy (initial - final).
- The work done by friction force is also equal to the product of friction force and distance.
- Therefore, we can write the equation:
Work done by friction force = Friction force * distance
(Friction force) * (distance) = (1/2) * (mass of block) * (initial velocity of block)^2
- Plugging in the given values:
(7.2N) * (distance) = (1/2) * (13kg) * (v_b)^2
- Rearranging the equation to solve for distance:
distance = [(1/2) * (13kg) * (v_b)^2] / (7.2N)
- Plugging in the initial velocity of the block, v_b, which is 0.459 m/s:
distance = [(1/2) * (13kg) * (0.459 m/s)^2] / (7.2N)
Calculating this, we find that the block will slide approximately 0.145 meters (or 14.5 cm) on the ice before stopping.