Recent content by Meezus

Ok. Is what I have done correct? I don't see where the rebound part comes into play?

Momentum after collision for carriage A: 0.5 x 2 = 1 Momentum after collision for carriage B:1 x ? = ? Total Momentum after collision: 3 If the total momentum stays the same the velocity for carriage B must be 2.

I thought because momentum is conserved the total momentum is the same after?

Like when you throw a ball at a wall and it comes back. It "rebounds" off wall.

Homework Statement A trolley of mass 0.5 kg rolls along a level, frictionless ramp at a velocity of 6.0 m s-1. It collides with a second trolley of mass 1.0 kg which is initially at rest. The first trolley rebounds with a speed of 2.0 m s-1 . a) Stating clearly any principles involved, find...

7200 kg.m/s?

a x t = change in V. 2 x 6 = 12. p=m x v 12 x 600 = 7200?

acceleration = change in V / t?

Homework Statement A vehicle has a mass of 600 kg. Its engine provides a forward force of 1500 N, but the motion is resisted by a constant frictional force of 300 N. Calculate a) the acceleration of the vehicle, b) its momentum 6.0 s after starting to move. Homework Equations F=ma a=f/m...

Ok thanks, I thought maybe gravity would have some impact here.

Nope, that's the whole question.

Homework Statement A 56 g tennis ball is accelerated at 1000 m s-2 to reach its service velocity. Calculate the average resultant force required to accelerate the ball. Homework Equations F = MA The Attempt at a Solution Mass = 0.56g = 0.056kg Acceleration = 1000m/s2 0.056 x 1000 = 56N...

I'm not sure how it should be expressed. -3-x = -(x+3). Should it be 1 - (3+x)? (IGNORE) Its (-3)(-x) so it should be 3x because its -3 * -x. sorry

1-3-x + 6-x2 - 10-x3

ooohh it all becomes negative? I see! So its 1-3x - 6x2?