You would need an impulse of 30,000 N⋅s to stop a 1,500 kg car traveling at 20 m/s.
How do you calculate impulse?
Impulse is calculated as the change in momentum, J = Δp, or as the product of average force and time, J = F⋅Δt.
Here's the thing: you can measure it two ways. Multiply the net force acting on an object by how long that force lasts. Or, if you know the object's mass and how much its velocity changes, use J = m⋅Δv. Take a 2 kg object slowing from 5 m/s to rest in 0.4 seconds — its impulse is (2 kg)×(−5 m/s) = −10 N⋅s. The force applied? (−10 N⋅s)/(0.4 s) = −25 N. Simple as that.
What is the impulse necessary to stop a 1500?
The impulse necessary to stop a 1,500 kg object moving at 25 m/s is 37,500 N⋅s.
That's J = m⋅Δv = (1,500 kg)×(25 m/s) = 37,500 N⋅s. Now, if the speed were 20 m/s instead — which is what you'd typically see in real-world car safety tests NHTSA — the required impulse drops to 30,000 N⋅s.
What is the momentum of a 1500 kg car traveling at 6 m/s?
The momentum of a 1,500 kg car traveling at 6 m/s is 9,000 kg⋅m/s.
Momentum is just mass times velocity: p = m⋅v. So 1,500 kg × 6 m/s = 9,000 kg⋅m/s. That tells you how tough it'd be to stop the car — imagine trying to halt a heavy bowling ball rolling slowly. It's still got plenty of oomph behind it.
How fast will a 1200 kg car be traveling when its momentum is 2400 kg/m/s?
The car will be traveling at 2 m/s when its momentum is 2,400 kg⋅m/s.
Plug the numbers into v = p/m: 2,400 kg⋅m/s ÷ 1,200 kg = 2 m/s. That's barely faster than a brisk walk. Try stopping a car moving that slowly without brakes — not easy, right?
What does area under force time curve gives us?
The area under a force-time curve gives the impulse delivered to an object.
This matters a lot in crash testing. A big force over a short time — like an airbag deploying — can deliver the same impulse as a small force over a longer time, like seatbelts. The effects on passenger safety differ wildly though IIHS.
How do you calculate the impulse needed to stop an object?
The impulse needed to stop an object is equal to its initial momentum, J = m⋅v.
To bring an object to a complete stop, its final momentum must hit zero. So impulse equals the negative of its initial momentum: J = Δp = p_final − p_initial = 0 − (m⋅v) = −m⋅v. The negative sign just means the force opposes the motion. Engineers use this to design crumple zones in cars — they absorb the right amount of impulse during a crash.
What is the impulse delivered to the ball?
The impulse delivered to the ball equals its change in momentum: Δp = m⋅(v_final − v_initial).
Say a 0.2 kg tennis ball hits the ground at 10 m/s downward and rebounds at 8 m/s upward. The change in velocity is 18 m/s, so impulse is 0.2 kg × 18 m/s = 3.6 N⋅s upward. That's the force the floor exerts on the ball during contact.
What is the symbol for impulse?
The symbol for impulse is J, and it’s equal to the change in momentum, Δp.
You'll also see impulse written as the integral of force over time: J = ∫ F dt. Physicists toss around J and Δp like they're interchangeable — and they are, thanks to the impulse-momentum theorem linking force, time, and change in motion.
How fast is a 1.50 kg ball moving if it has a momentum of 4.50 kg/m s?
The ball is moving at 3 m/s.
Just divide momentum by mass: v = p/m = 4.50 kg⋅m/s ÷ 1.50 kg = 3 m/s. Picture a basketball rolling at a brisk walking speed — that's roughly how fast it's moving.
How do you calculate change in momentum?
Change in momentum is calculated as Δp = m⋅(v_final − v_initial).
This is the impulse. Take a hockey puck (0.17 kg) that speeds up from 10 m/s to 25 m/s. Its Δp = 0.17 kg × (25 − 10) = 2.55 kg⋅m/s. That change could come from a stick applying force in a fraction of a second.
Which situation will produce the greatest change of momentum?
A larger force acting over a short time produces the greatest change in momentum when the total impulse is the same.
Think baseball bat hitting a ball. A hard swing (high force) over brief contact transfers the same impulse as a soft swing over longer time. But the high-force hit causes a rapid change, sending the ball flying.
What is the momentum of a 2000 kg car traveling at 20 m/s?
The momentum is 40,000 kg⋅m/s.
Calculate p = m⋅v = 2,000 kg × 20 m/s = 40,000 kg⋅m/s. That's a ton of momentum — enough to make stopping the car require serious force, like braking over several car lengths.
What is the momentum of a 1300 kg car traveling with a speed of 28 m/s?
The momentum is 36,400 kg⋅m/s.
p = 1,300 kg × 28 m/s = 36,400 kg⋅m/s. At highway speeds, even a small car packs serious momentum — that's why seatbelts and airbags matter so much.
What is the change in momentum of a 3 kg object accelerating from rest to 12 m/s?
The change in momentum is 36 kg⋅m/s.
Δp = m⋅Δv = 3 kg × (12 − 0) = 36 kg⋅m/s. That's the impulse the accelerating force must provide — whether from a rocket engine, a gust of wind, or a push.
What is the momentum of a 1500 kg car traveling at 6 m s?
The momentum of a 1,500 kg car traveling at 6 m/s is 9,000 kg m/s.
How fast will a 1200 kg car be traveling when its momentum is 2400 kg/m s?
The car will be traveling at 2 m/s when its momentum is 2,400 kg⋅m/s.
What is the momentum of a 2000 kg car traveling at 20 ms?
The momentum is 40,000 kg⋅m/s.
What is the momentum of a 1300 kg car traveling with a speed of 28 m s?
The momentum is 36,400 kg⋅m/s.
What is the change in momentum of a 3 kg object accelerating from rest to 12 m s?
The change in momentum is 36 kg⋅m/s.
