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M1 Impulse/Momentum
Hello again. 
Is anybody able to show be how to do the following?
A ball of mass 0.4kg is dropped from a height of 2.5m. After hitting the ground it rises to a height of 1.8m. Find (a) the speed with which the the ball rebounds from the ground and (b) the impulse the ball receives from the ground.
Thanks.
Is anybody able to show be how to do the following?A ball of mass 0.4kg is dropped from a height of 2.5m. After hitting the ground it rises to a height of 1.8m. Find (a) the speed with which the the ball rebounds from the ground and (b) the impulse the ball receives from the ground.
Thanks.
Sorry. Yes, got that one sorted. Stuck on the following though (and the search function isn't working as usual)... so if you're still in the mood for thinking... 
Particle A of mass 1kg is at rest 0.2m from the edge of a smooth horizontal 0.8m high table. It is connected by a light inextensible string of length 0.7m to a particle B of mass 0.5kg. Particle B is initially at rest at the edge of the table closest to A but then falls off. Assuming B's initial horizontal velocity to be zero find the speed with which A begins to move.
Particle A of mass 1kg is at rest 0.2m from the edge of a smooth horizontal 0.8m high table. It is connected by a light inextensible string of length 0.7m to a particle B of mass 0.5kg. Particle B is initially at rest at the edge of the table closest to A but then falls off. Assuming B's initial horizontal velocity to be zero find the speed with which A begins to move.
I'm not sure about this one, not sure if I'm picturing it right, but see what you think.
First, I assume A starts to move when B has fallen 0.5m below the edge of the table.
Use v^2 = u^2 + 2as to find the speed of B at this point (3.13 m/s)
So, if this is a momentum questions, I assume we now use conservation of momentum? Momentum of B just before A starts to move is 0.5 x 3.13 = 1.56.
So momentum of total system once A also starts to move is the same as this, so the speed of the two together is momentum / mass = 1.56 /1.5
Does that sound right?
First, I assume A starts to move when B has fallen 0.5m below the edge of the table.
Use v^2 = u^2 + 2as to find the speed of B at this point (3.13 m/s)
So, if this is a momentum questions, I assume we now use conservation of momentum? Momentum of B just before A starts to move is 0.5 x 3.13 = 1.56.
So momentum of total system once A also starts to move is the same as this, so the speed of the two together is momentum / mass = 1.56 /1.5
Does that sound right?
Particle A of mass 1kg is at rest 0.2m from the edge of a smooth horizontal 0.8m high table. It is connected by a light inextensible string of length 0.7m to a particle B of mass 0.5kg. Particle B is initially at rest at the edge of the table closest to A but then falls off. Assuming B's initial horizontal velocity to be zero find the speed with which A begins to move.
to make it much easier to solve this kind of problems , you should analyse what you have.
1.they both move in the same direction so both velocitys are +
2.the speed of A when the string becomes taut , is the initial speed.
now standard suvat protocool:
u = 0
s = 0.7-0.2 = 0.5
a = 9.8
v = ?
v2 = u2 + 2(as) (sub values)
v2 = 2(0.5 * 9.8) = 9.8
v = sqr(9.8)
now standard conversion of momentum protocool
m1 = 1kg
m2 = 0.5kg
u1 = 0
u2 = sqr(9.8)
v1 = v2
all speeds are in the same direction and both final speeds are equal.
m1u1 + m2u2 = m1v + m2v (sub values)
1*0 + 0.5*sqr(9.8) = 1*v + 0.5*v (simplify)
1.5v = 0.5*sqr(9.8)
v = (0.5*sqr(9.8)) / 1.5 = 1.04 (3 s.f)
hope this helps.
to make it much easier to solve this kind of problems , you should analyse what you have.
1.they both move in the same direction so both velocitys are +
2.the speed of A when the string becomes taut , is the initial speed.
now standard suvat protocool:
u = 0
s = 0.7-0.2 = 0.5
a = 9.8
v = ?
v2 = u2 + 2(as) (sub values)
v2 = 2(0.5 * 9.8) = 9.8
v = sqr(9.8)
now standard conversion of momentum protocool
m1 = 1kg
m2 = 0.5kg
u1 = 0
u2 = sqr(9.8)
v1 = v2
all speeds are in the same direction and both final speeds are equal.
m1u1 + m2u2 = m1v + m2v (sub values)
1*0 + 0.5*sqr(9.8) = 1*v + 0.5*v (simplify)
1.5v = 0.5*sqr(9.8)
v = (0.5*sqr(9.8)) / 1.5 = 1.04 (3 s.f)
hope this helps.
Original post by Codefusion
Particle A of mass 1kg is at rest 0.2m from the edge of a smooth horizontal 0.8m high table. It is connected by a light inextensible string of length 0.7m to a particle B of mass 0.5kg. Particle B is initially at rest at the edge of the table closest to A but then falls off. Assuming B's initial horizontal velocity to be zero find the speed with which A begins to move.
to make it much easier to solve this kind of problems , you should analyse what you have.
1.they both move in the same direction so both velocitys are +
2.the speed of A when the string becomes taut , is the initial speed.
now standard suvat protocool:
u = 0
s = 0.7-0.2 = 0.5
a = 9.8
v = ?
v2 = u2 + 2(as) (sub values)
v2 = 2(0.5 * 9.8) = 9.8
v = sqr(9.8)
now standard conversion of momentum protocool
m1 = 1kg
m2 = 0.5kg
u1 = 0
u2 = sqr(9.8)
v1 = v2
all speeds are in the same direction and both final speeds are equal.
m1u1 + m2u2 = m1v + m2v (sub values)
1*0 + 0.5*sqr(9.8) = 1*v + 0.5*v (simplify)
1.5v = 0.5*sqr(9.8)
v = (0.5*sqr(9.8)) / 1.5 = 1.04 (3 s.f)
hope this helps.
to make it much easier to solve this kind of problems , you should analyse what you have.
1.they both move in the same direction so both velocitys are +
2.the speed of A when the string becomes taut , is the initial speed.
now standard suvat protocool:
u = 0
s = 0.7-0.2 = 0.5
a = 9.8
v = ?
v2 = u2 + 2(as) (sub values)
v2 = 2(0.5 * 9.8) = 9.8
v = sqr(9.8)
now standard conversion of momentum protocool
m1 = 1kg
m2 = 0.5kg
u1 = 0
u2 = sqr(9.8)
v1 = v2
all speeds are in the same direction and both final speeds are equal.
m1u1 + m2u2 = m1v + m2v (sub values)
1*0 + 0.5*sqr(9.8) = 1*v + 0.5*v (simplify)
1.5v = 0.5*sqr(9.8)
v = (0.5*sqr(9.8)) / 1.5 = 1.04 (3 s.f)
hope this helps.
Hey why did you "0.7-0.5" to get s?
and also how did the masses change on the second part of the equation?
(edited 11 years ago)
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