A student throws a .1kg ball at a wall. The ball hits the wall perpendicularly with a speed of 5m/s & then bounces back w/ a new speed of 4m/s. What impulse is given to the ball by the wall?
I used deltaP= m deltaV so deltaP= .1*(-1) to get -.1N-s.
I got it wrong! Can someone tell me why?
Thank you :)
the change in velocity (delta v) is
... 9 m/s
5 m/s in one direction, then 4 m/s in the opposite direction
Oh... So essentially I'm saying 5-(-4) which is 5+4? that makes sense... Thank you :)
To correctly calculate the impulse given to the ball by the wall, you need to use the principle of conservation of momentum. According to this principle, the total momentum before and after the collision should be the same.
In this case, let's consider the wall as the system and analyze the collision. Before the collision, the ball is moving towards the wall with a speed of 5 m/s, so its momentum is given by:
P1 = m * v1 = 0.1 kg * 5 m/s = 0.5 kg·m/s
Now, after the collision, the ball bounces back with a speed of 4 m/s. As the direction of motion has changed, we need to assign a negative sign to the velocity. Therefore, the final momentum of the ball is given by:
P2 = m * v2 = 0.1 kg * (-4 m/s) = -0.4 kg·m/s
Now, to find the impulse given to the ball by the wall, we can calculate the change in momentum:
ΔP = P2 - P1 = (-0.4 kg·m/s) - (0.5 kg·m/s) = -0.9 kg·m/s
The impulse given to the ball by the wall is -0.9 kg·m/s.
It seems you made a mistake while calculating the change in momentum. You should subtract the initial momentum (P1) from the final momentum (P2) to find the change in momentum (ΔP), rather than multiplying the mass by the change in velocity.