a 1 kg dart moving horizntally at 10 m/s sticks to a wood block of mass 9 kg which slides across a friction free level surface. what is the speed of the block and dart after collision?
Conservation of momentum applies
1kg*10=(1+9)kg*V
500000
wrong answer
To find the speed of the block and dart after the collision, we can use the principle of conservation of momentum. According to this principle, the total momentum before the collision is equal to the total momentum after the collision.
The momentum (p) of an object is given by the product of its mass (m) and velocity (v), so we can express the momentum as p = mv.
Before the collision, the dart has a mass (m1) of 1 kg and a velocity (v1) of 10 m/s. The wood block has a mass (m2) of 9 kg and is initially at rest, so its velocity (v2) is 0 m/s.
The total momentum before the collision is therefore:
p_initial = (m1 * v1) + (m2 * v2)
After the collision, the dart becomes part of the wood block, so both objects move together as a single system. Let's assume the final velocity of the block and dart after the collision is v_final.
The total momentum after the collision is therefore:
p_final = (m1 + m2) * v_final
Since momentum is conserved, we can set the initial momentum equal to the final momentum:
p_initial = p_final
(m1 * v1) + (m2 * v2) = (m1 + m2) * v_final
Plugging in the given values:
(1 kg * 10 m/s) + (9 kg * 0 m/s) = (1 kg + 9 kg) * v_final
10 kg * m/s = 10 kg * v_final
Dividing both sides by 10 kg gives:
m/s = v_final
Therefore, the speed of the block and dart after the collision is 1 m/s.