Question

A 21.0 kg shopping cart is moving with a velocity of 4.0 m / s. It strikes a 7.0 kg box that is initially at rest. They stick together and continue moving at a new velocity. Assume that friction is negligible.
What was the momentum of the shopping cart before the collision? 168 kg dot m/s 84.0 kg dot m/s 44.0 kg dot m/s 0 kg dot m/s

What was the momentum of the box before the collision? 84.0 kg dot m/s 0 kg dot m/s 168 kg dot m/s 44.0 kg dot m/s

What is the velocity of the combined shopping cart/box wreckage after the collision? 4.0 m/s 7.64 m/s 2.6 m/s 0 m/s

Answer 1

Explanation:

Given that,

Mass of the shopping cart, $m_s=21\ kg$

Initial speed of the shopping cart, $u_s=4\ m/s$

Mass of the box, $m_b=7\ kg$

Initial speed of the box, $u_b=0$ (at rest)

They stick together and continue moving at a new velocity. It is a case of inelastic collision.

(a) The momentum of the shopping cart before the collision is given by :

$p_s=m_s\times u_s\\\\p_s=21\times 4\\\\p_s=84\ kg-m/s$

(b) The momentum of the box before the collision is given by :

$p_b=m_b\times u_b\\\\p_b=7\times 0\\\\p_b=0$

(c) The velocity of the combined shopping cart/box wreckage after the collision is given by using the conservation of momentum as :

$m_su_s+m_bu_b=(m_s+m_b)V\\\\V=\dfrac{m_su_s+m_bu_b}{(m_s+m_b)}\\\\V=\dfrac{21\times 4+0}{(21+7)}\\\\V=3\ m/s$

Hence, this is the required solution.