truck A of mass 2,5 x 10(4) kg approaches a second truck B of mass 2 x 10(4) kg with a velocity of 12 m.s -1 and collides with it. After collision the two trucks move apart with equal but opposite momenta. Calculate what the velocity of B was before the collision
To calculate the velocity of truck B before the collision, we can make use of the principle of conservation of momentum.
The principle of conservation of momentum states that the total momentum before a collision is equal to the total momentum after the collision, assuming there are no external forces acting on the system.
Initially, truck A is approaching truck B with a velocity of 12 m/s, and we assume truck B is initially at rest (velocity of 0 m/s). After the collision, both trucks move apart with equal but opposite momenta.
Let's denote the velocity of truck B after the collision as v_B' (final velocity of truck B) and v_A' (final velocity of truck A).
According to the principle of conservation of momentum:
(mass of truck A) × (initial velocity of truck A) + (mass of truck B) × (initial velocity of truck B)
= (mass of truck A) × (final velocity of truck A) + (mass of truck B) × (final velocity of truck B)
Substituting the given values:
(2.5 × 10^4 kg) × (12 m/s) + (2 × 10^4 kg) × (0 m/s)
= (2.5 × 10^4 kg) × (v_A') + (2 × 10^4 kg) × (v_B')
Simplifying this equation, we can solve for v_B':
(2.5 × 10^4 kg) × (12 m/s) = (2.5 × 10^4 kg) × (v_A') + (2 × 10^4 kg) × (v_B')
300,000 kg·m/s = (2.5 × 10^4 kg) × (v_A') + (2 × 10^4 kg) × (v_B')
Since the two trucks move apart with equal but opposite momentum, we know that (2.5 × 10^4 kg) × (v_A') = (2 × 10^4 kg) × (-v_B'), meaning the magnitudes of their momenta are equal.
Substituting this relationship into the equation above:
300,000 kg·m/s = (2.5 × 10^4 kg) × (v_A') + (-2 × 10^4 kg) × (v_A')
Rearranging the equation to solve for v_A', we get:
(2.5 × 10^4 kg - 2 × 10^4 kg) × (v_A') = 300,000 kg·m/s
(0.5 × 10^4 kg) × (v_A') = 300,000 kg·m/s
v_A' = (300,000 kg·m/s) / (0.5 × 10^4 kg)
v_A' = 6000 m/s
Therefore, the velocity of truck B before the collision is 6000 m/s.