A 45 kg student is riding on a 7kg scateboard with a velocity of +4m/s. The student jumps of the cart with a velocity of -1m/s. Find the velocity of the scateboard after the student jumped off.
conserve momentum.
(45+7)*4 = 45(-1) + 7v
A student
To find the velocity of the skateboard after the student jumped off, we can apply the principle of conservation of momentum.
The formula for momentum is:
momentum = mass × velocity
According to the law of conservation of momentum, the total momentum before the student jumps off must be equal to the total momentum after the student jumps off.
Before the student jumps off, the total momentum is given by the sum of the momentum of the student and the momentum of the skateboard:
Initial momentum = (mass of student × velocity of student) + (mass of skateboard × velocity of skateboard)
The mass of the student is 45 kg, the velocity of the student is -1 m/s, the mass of the skateboard is 7 kg, and the velocity of the skateboard is +4 m/s.
Initial momentum = (45 kg × -1 m/s) + (7 kg × 4 m/s)
Initial momentum = -45 kg·m/s + 28 kg·m/s
Initial momentum = -17 kg·m/s
After the student jumps off, the total momentum is equal to the momentum of the skateboard. Let's denote the final velocity of the skateboard as 'v'.
Final momentum = mass of skateboard × final velocity of skateboard
Final momentum = 7 kg × v
Since the total momentum before and after the student jumps off are equal, we can set them equal to each other:
Initial momentum = Final momentum
-17 kg·m/s = 7 kg × v
To solve for 'v', we can rearrange the equation:
v = (-17 kg·m/s) / 7 kg
v ≈ -2.4 m/s
Therefore, the velocity of the skateboard after the student jumped off is approximately -2.4 m/s.
To find the velocity of the skateboard after the student jumps off, we can use the law of conservation of momentum.
The law of conservation of momentum states that the total momentum of a system remains constant if no external forces act on it.
The momentum of an object is defined as the product of its mass and velocity. It is a vector quantity, which means it has both magnitude and direction.
Let's define the variables:
Mass of the student (m1) = 45 kg
Mass of the skateboard (m2) = 7 kg
Initial velocity of the skateboard (v2i) = +4 m/s
Final velocity of the student (v1f) = -1 m/s
Final velocity of the skateboard (v2f) = ?
According to the law of conservation of momentum, the initial momentum of the system (student + skateboard) is equal to the final momentum of the system.
The initial momentum (p_initial) is the sum of the individual momenta before the student jumps off:
p_initial = m1 * v1_initial + m2 * v2_initial
Since the student is riding the skateboard without external forces acting on the system, the initial momentum is:
p_initial = (m1 + m2) * v_initial
The final momentum (p_final) is the sum of the individual momenta after the student jumps off:
p_final = m1 * v1_final + m2 * v2_final
Since the total momentum is conserved, we have:
p_initial = p_final
Therefore:
(m1 + m2) * v_initial = m1 * v1_final + m2 * v2_final
Substituting the given values:
(45 kg + 7 kg) * 4 m/s = 45 kg * (-1 m/s) + 7 kg * v2_final
Simplifying:
52 kg * 4 m/s = -45 kg * 1 m/s + 7 kg * v2_final
208 kg*m/s = -45 kg*m/s + 7 kg * v2_final
Adding 45 kg*m/s to both sides:
208 kg*m/s + 45 kg*m/s = 7 kg * v2_final
253 kg*m/s = 7 kg * v2_final
Dividing both sides by 7 kg:
v2_final = 253 kg*m/s / 7 kg
v2_final = 36.14 m/s
Therefore, the velocity of the skateboard after the student jumps off is approximately +36.14 m/s.