1.)A 0.5 kg hockey puck moving at 35 m/s hits a straw bale, stopping in 1 s.
a) What impulse is delivered to the ball?
b) What force is exerted on the puck?
2.) A racing car with a mass of 1400 kg hits a slick spot and crashes head-on into a concrete wall at 50 m/s, coming to a halt in 0.8 s.
a) What is the change in momentum?
b) What is the impulse
c)What force is exerted on the car?
3.) An ambulance weighing 3000 kg comes racing to the rescue, hits the same slick spot, and then collides with a padded part of the wall at 50 m/s, coming to a halt in 2 s.
a) What is the change in momentum?
b) What is the impulse
c)What force is exerted on the ambulance?
d) How this answer differ from the problem above??
>>FORMULAS<<
P=w/t
P= energy/time
P=mad/t
P= Fd/t
help please!! :)
To solve these problems, we will use the formulas related to impulse, force, momentum, and energy. The formulas we will need are:
- Impulse (J): Impulse is defined as the change in momentum of an object and is given by the formula J = Δp = mΔv. Where J is impulse, Δp is the change in momentum, m is the mass of the object, and Δv is the change in velocity.
- Force (F): The force exerted on an object during a collision or interaction can be calculated using the formula F = mΔv / Δt, where F is the force, m is the mass, Δv is the change in velocity, and Δt is the time interval.
- Momentum (p): Momentum is the product of an object's mass and velocity and is calculated using the formula p = mv, where p is momentum, m is mass, and v is velocity.
- Power (P): Power is the rate at which work is done, or energy is transferred, and it is calculated using the formula P = W / t, where P is the power, W is the work done, and t is the time taken.
Now let's solve the given problems step-by-step:
1) Hockey Puck Problem:
a) To find the impulse delivered to the puck, we need to calculate the change in momentum. Given that the mass (m) of the puck is 0.5 kg and the initial velocity (v) is 35 m/s, the change in velocity (Δv) is 0 m/s since the puck stops. Therefore, the impulse (J) delivered to the puck is J = mΔv = (0.5 kg)(0 m/s) = 0 kg*m/s.
b) To find the force exerted on the puck, we can use the formula F = mΔv / Δt. Since the puck stops in 1 second, the time interval (Δt) is 1 second. Substituting the values, we get F = (0.5 kg)(0 m/s) / 1 s = 0 N. Therefore, the force exerted on the puck is 0 N.
2) Racing Car Problem:
a) To find the change in momentum, we multiply the mass (m) of the car (1400 kg) by the change in velocity (Δv). Given that the initial velocity (v) is 50 m/s and the car comes to a halt, the change in velocity is -50 m/s (negative sign indicates opposite direction). Therefore, the change in momentum (Δp) is Δp = mΔv = (1400 kg)(-50 m/s) = -70,000 kg*m/s.
b) The impulse (J) is the same as the change in momentum. So, J = -70,000 kg*m/s.
c) To find the force exerted on the car, we can use the formula F = mΔv / Δt. Given that the time interval (Δt) is 0.8 seconds, substituting the values, we get F = (1400 kg)(-50 m/s) / 0.8 s = -87,500 N (negative sign indicates opposite direction). Therefore, the force exerted on the car is -87,500 N.
3) Ambulance Problem:
a) To find the change in momentum, we multiply the mass (m) of the ambulance (3000 kg) by the change in velocity (Δv). Given that the initial velocity (v) is 50 m/s and the ambulance comes to a halt, the change in velocity is -50 m/s. The change in momentum (Δp) is Δp = mΔv = (3000 kg)(-50 m/s) = -150,000 kg*m/s.
b) The impulse (J) is the same as the change in momentum. So, J = -150,000 kg*m/s.
c) To find the force exerted on the ambulance, we use the formula F = mΔv / Δt. Given that the time interval (Δt) is 2 seconds, substituting the values, we get F = (3000 kg)(-50 m/s) / 2 s = -75,000 N. Therefore, the force exerted on the ambulance is -75,000 N.
d) The answer in this problem differs from the racing car problem in terms of the force exerted on the ambulance (-75,000 N) compared to the force exerted on the car (-87,500 N). This difference can be explained by the difference in the time intervals (2 seconds for the ambulance and 0.8 seconds for the car) during which the velocity of the objects changes. A longer time interval spreads the change in momentum over a greater time period, resulting in a lesser force exerted.