a car of mass 600 kg is moving at a speed of 20 m/s. it collides with a stationary car of mass 900 kg. if the first car bounces back at 4 m/s,at what speed does the second car move after the collision?
12000=2400+900v
v=16m/s
To solve this problem, we can use the principle of conservation of momentum. The momentum before the collision must equal the momentum after the collision.
Let's denote the initial velocity of the first car as v1i, the final velocity of the first car as v1f, the initial velocity of the second car as v2i, and the final velocity of the second car as v2f.
The momentum before the collision can be calculated as the sum of the individual momenta of the two cars. Therefore:
Initial momentum before the collision = (mass of first car * initial velocity of first car) + (mass of second car * initial velocity of second car)
= (600 kg * 20 m/s) + (900 kg * 0 m/s)
= 12,000 kg⋅m/s
According to the conservation of momentum, this value must be equal to the momentum after the collision, which is given only in terms of the final velocity of the first car:
Final momentum after the collision = (mass of first car * final velocity of first car) + (mass of second car * final velocity of second car)
= (600 kg * -4 m/s) + (900 kg * v2f)
= -2,400 kg⋅m/s + 900 kg⋅v2f
Since momentum is conserved, we can equate the two expressions:
12,000 kg⋅m/s = -2,400 kg⋅m/s + 900 kg⋅v2f
Rearranging the equation, we get:
14,400 kg⋅m/s = 900 kg⋅v2f
Dividing both sides of the equation by 900 kg, we find:
v2f = 16 m/s
Therefore, after the collision, the second car moves with a speed of 16 m/s.
To find the speed of the second car 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 of an object is calculated by multiplying its mass by its velocity. Mathematically, momentum (p) is given by:
p = m * v
Where:
p = momentum
m = mass
v = velocity
Let's denote the first car as car 1 (mass = 600 kg, initial velocity = 20 m/s, final velocity = -4 m/s) and the second car as car 2 (mass = 900 kg, final velocity = ?).
Initially, the momentum of car 1 is:
p1_initial = m1 * v1_initial
= 600 kg * 20 m/s
= 12,000 kg·m/s
Initially, the momentum of car 2 (which is stationary) is:
p2_initial = 0, as the initial velocity is 0.
After the collision, the momentum of car 1 is:
p1_final = m1 * v1_final
= 600 kg * (-4 m/s) // Since the car is bouncing back, the velocity is negative.
= -2,400 kg·m/s
According to the principle of conservation of momentum, the total momentum before the collision (p1_initial + p2_initial) should be equal to the total momentum after the collision (p1_final + p2_final). Thus:
p1_initial + p2_initial = p1_final + p2_final
Substituting the values we know:
12,000 kg·m/s + 0 kg·m/s = -2,400 kg·m/s + p2_final
Simplifying the equation, we find that:
p2_final = 14,400 kg·m/s
Now, we can calculate the final velocity (v2_final) of car 2 after the collision by rearranging the momentum equation:
p2_final = m2 * v2_final
Rearranging the equation to find v2_final, we have:
v2_final = p2_final / m2
Substituting the known values:
v2_final = 14,400 kg·m/s / 900 kg
= 16 m/s
Therefore, after the collision, the second car moves with a speed of 16 m/s.